CHAPTER 1 PRELIMINARIES 1.1 REAL NUMBERS AND THE REAL LINE 1. Executing long division, ^ = 0.1, 0.2, | = 0.3, | 0.8, | 0.9 2. Executing long division, yj 0.09, fy 0.18, f\ 27 — 0.81, ii 0.99 NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6. a) NNT. 5 is a counter example. b) NT. 2<x<6 ^2-2<x-2<6-2 =4>0<x-2<2. c) NT. 2 < x < 6 => 2/2 < x/2 < 6/2 => 1 < x < 3. d) NT. 2 < x < 6 => 1/2 > 1/x > 1/6 => 1/6 < 1/x < 1/2. e) NT. 2 < x < 6 => 1/2 > 1/x > 1/6 => 1/6 < 1/x < 1/2 -- f) NT. 2<x<6 => x<6=>(x-4)<2and2<x<6 The pair of inequalities (x — 4) < 2 and — (x — 4) < 2 g) NT. 2 < x < 6 => -2 > -x > -6 =4> -6 < -x < -2. But -2 < 2. So -6 < -x < -2 < 2 or -6 < -x < 2. h) NT. 2<x<6 => -1(2)>-I(x)<-1(6) => -6 < -x < -2 6(1/6) <6(l/x)< 6(1/2) > x>2 => -x< -2 => > I x — 4 I < 2. > 1 < 6/x < 3. -x + 4<2 =4> (x-4)<2. 4. NT = necessarily true, NNT = Not necessarily true. Given: — l<y — 5<1. a) NT. -1 <y - 5 < 1 =>-l+5<y-5 + 5<l+5 => 4 < y < 6. b) NNT. y = 5 is a counter example. (Actually, never true given that 4 < y < 6) c) NT. From a), — 1 < y - 5 < 1, =>4<y<6=^>y>4. d) NT. From a), - 1 < y - 5 < 1, =>4<y<6=>y<6. e) NT. — 1 <y — 5 < 1 =>-l + l<y-5+l<l + l =>0<y-4<2. f) NT. -I<y-5<1 => (l/2)(-l +5)<(l/2)(y- 5 + 5)< (l/2)(l+5) =>■ 2 < y/2 < 3. g) NT. From a), 4 < y < 6 =>■ 1/4 > 1/y > 1/6 => 1/6 < 1/y < 1/4. h) NT. — 1 < y — 5 < 1 => y - 5 > -1 => y>4 =>• -y < -4 => -y + 5 < 1 => -(y-5)<l. Also, — l<y — 5<1 => y — 5 < 1. The pair of inequalities — (y — 5) < 1 and (y — 5) < 1 =>• | y — 5 | < 1. 5. -2x > 4 => x < -2 6. 8 - 3x > 5 => -3x > -3 => x < 1 -> x 7. 5x - 3 < 7 - 3x => 8x < 10 => x < 3(2 - x) > 2(3 + x) =>■ 6 - 3x > 6 + 2x =4> 0>5x => 0>x 9. 2x-i>7x+|^>-i-|>5x => l(-jf)>xor-i>i 10. ^ < ^ => 12 - 2x < 12x - 16 =4> 28 < 14x => 2 < x k)2 2 Chapter 1 Preliminaries 11. | (x - 2)< i (x - 6) =4> 12(x - 2) < 5(x - 6) =4> 12x - 24 < 5x - 30 =4> 7x < -6 or x < - | 12. x±5 < i2±3x ^ _ (4x + 20) < 24 + 6x > -44 < lOx =*• - # < x -22/5 13. y = 3 or y = — 3 14. y - 3 = 7 or y - 3 = -7 =4> y = 10 or y = -4 15. 2t + 5 = 4or2t + 5 = -4 => 2t=-lor2t=-9 =5> t=-| or t 16. 1 - t = 1 or 1 - t = - 1 => -t = or -t = -2 =>■ t = or t = 2 17. 8 - 3s = | or 8 - 3s = - § =4> -3s = - \ or -3s 25 2 7 25 S = 6 OT S = -6 18. § - 1 = 1 or | - 1 = -1 => |=2 or |=0 => s = 4ors = 19. —2 < x < 2; solution interval (—2, 2) 20. —2 < x < 2; solution interval [—2, 2] -> x 21. — 3 < t— 1 < 3 => -2 < t < 4; solution interval [-2, 4] 22. — 1 < t + 2 < 1 => —3 < t < — 1; solution interval (—3, — 1) -»t 23. -4 < 3y - 7 < 4 =^ 3 < 3y < 11 => 1 < y < ^ ; solution interval ( 1 , 11/3 24. -1 < 2y + 5 < 1 =>• -6 < 2y < -4 =4> solution interval (—3, —2) -3<y < -2; 25. -1 < | - 1 < 1 =>• < f < 2 =^ < z < 10; solution interval [0, 10] 26. -2 < f - 1 < 2 => -1 < f < 3 => 3z | < z < 2; solution interval [— 1,2] -2/3 27. I<3 1 < i x ^ 2 7 - < -1 < -5 2x2 7 15 2 "^ x -^ 2 | < x < | ; solution interval (1,1) 28. -3 < ? - 4 < 3 =>• 1 < ? < 7 ^ 1 > | > i =4> 2 > x > | => 2^7 | < x < 2; solution interval (1,2) 2/7 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 1 . 1 Real Numbers and the Real Line 29. 2s > 4 or -2s > 4 => s > 2 or s < -2; solution intervals (— oo, —2] U [2, oo) 30. s + 3 > \ or -(s + 3) > \ => s > - \ or -s > \ => s>— | or s < — | ; solution intervals -00,-^jU |,oo) -7/2 -5/2 31. 1 - x > 1 or -(1 - x) > 1 => -x > or x > 2 =>• x<0orx>2; solution intervals (— oo, 0) U (2, oo) 32. 2 - 3x > 5 or -(2 - 3x) > 5 => -3x > 3 or 3x > 7 =4> x<— lorx>|; solution intervals (— oo, — 1) U (|, oo) 7/3 33. !±1 > 1 or 'r+n > 1 =>• r+ 1 > 2orr+ 1 < -2 =>• r > 1 or r < —3; solution intervals (— oo, —3] U [1, oo) 34. |-1> |or-(f -1) > \ =^ ^>|or — ^>— | => r > | or r < 1 solution intervals (— oo, 1) U ( |, oo) 7/3 35. x 2 < 2 => |x| < i/2 => -a/2 < x < a/2; solution interval I — a/2, a/ 2 ] -a/2 a/2 36. 4 < x 2 => 2 < |x| =>• x > 2 or x < -2; solution interval (—00, —2] U [2, 00) -2 37. 4<x 2 <9 =*> 2<|x|<3 => 2<x<3or2<-x<3 =>■ 2 < x < 3 or -3 < x < -2; solution intervals (—3, —2) U (2, 3) 38. i < x 2 < \ \<\x\<\ =4> 5<x<iorj<-x<i =>■ 5 < x < 1 or solution intervals (■ < x < 5JU(f,f) 1/2 -1/3 1/3 1/2 39. (x - l) 2 < 4 => |x - 1| < 2 => -2 < x - 1 < 2 =>• — 1 < x < 3; solution interval (—1,3) 40. (x + 3) 2 <2 => |x + 3| < a/2 =>■ -a/2 <x + 3<\/2 or -3-a/2<x<-3 + a/2; solution interval I —3 — a/2, —3 + a/2 1 -3 -a/2 -3+ a/2 Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Chapter 1 Preliminaries 41. x 2 - x < => x — x+j< j => (x So the solution is the interval (0, 1) l) 2 <i 2 J v 4 1 < i 2^2 ■\ <x - 1 < | => 0<x<l. 42. x 2 - x - 2 > => x 2 X+ 4-4^ X 2—2 3 > | or -(x - i) > | =>• x > 2or x < -1. The solution interval is (— oo, — 1] U [2, oo) 43. True if a > 0; False if a < 0. 44. |x - 1| = 1 - x O |-(x - 1)| = 1 -x 44> 1 - x > <£► x < 1 45. (1) |a + b| = (a + b) or |a + b| = -(a + b); both squared equal (a + b) 2 (2) ab < |ab| = |a| |b| (3) |a| = a or |a| = —a, so |a| = a 2 ; likewise, |b| = b 2 (4) x 2 < y 2 implies v x 2 < y> or x < y for all nonnegative real numbers x and y. Let x = |a + b| and y = |a| + |b| so that |a + b| 2 < (|a| + |b|) 2 =» |a + b| < |a| + |b| . 46. If a > and b > 0, then ab > and |ab| = ab = |a| |b| . If a < and b < 0, then ab > and |ab| = ab = (-a)(-b) = |a| |b| . If a > and b < 0, then ab < and |ab| = -(ab) = (a)(-b) = |a| |b| . If a < and b > 0, then ab < and |ab| = -(ab) = (-a)(b) = |a| |b| . 47. -3 < x < 3 and x > - \ => - \ < x < 3. 48. Graph of |x| + |y| < 1 is the interior of "diamond-shaped" region. x| + |y|<l 49. Let S be a real number > and f(x) = 2x + 1 . Suppose that | x — 1 \<6. Then | x— 1 | < 6 => 2| x— 1 | < 26 | 2x - 2 | < 26 => | (2x + 1) - 3 | < 26 =4> | f(x) - f(l) | < 26 50. Let e > be any positive number and f(x) = 2x + 3. Suppose that | x — | < e/2. Then 2| x — | < e and | 2x + 3 -3 | < e. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) - f(0) | < e. 51. Consider: i) a > 0; ii) a < 0; iii) a = 0. i) For a > 0, | a | = a by definition. Now, a > => — a < 0. Let —a = b. By definition, | b | = — b. Since b = —a, | — a | = —(—a) = a and | a | = | — a | = a. ii) For a < 0, | a | = —a. Now, a < => — a > 0. Let —a = b. By definition, | b | = b and thus |— a| = —a. So again | a | = |— a|. iii) By definition | | = and since —0 = 0, | — | = 0. Thus, by i), ii), and iii) | a | = | — a | for any real number. Cop # (c) 1 Pearson Etation, k, publishing as Pearson iii-lslo Section 1.2 Lines, Circles and Parabolas 52. i) Prove | x | > =>■ x > a or x < — a for any positive number, a. For x > 0, | x | = x. | x | > a => x>a. Forx<0, |x| = — x. | x | > a =>■ — x > a => x < — a. ii) Prove x > a or x < — a =>• | x | > for any positive number, a. a > and x>a => |x| = x. So x > a =4> |x|>a. For a > 0, — a < and x < — a => x<0 => I x I = — x. So x < — a =4> — x > a =$■ I x I > a. 53. a) 1 = 1 => | 1 1 \ I ^ I b) nrr = a Mil b |b| 54. Prove S n = |a n | = |a| n for any real number a and any positive integer n. la 1 ! = |a| : = a, so Si is true. Now, assume that Sk = |a k | = |a| k is true form some positive integer k. Since la 1 ! = lal : and |a k | = tat k , we have |a k+1 | = |a k ■ a 1 ! = |a k ||a 1 | = lal k lal x = |a| k+1 . Thus, S k +i = la k+ll a| is also true. Thus by the Principle of Mathematical Induction, S n = | a n | = | a | n is true for all n positive integers. 1.2 LINES, CIRCLES, AND PARABOLAS 1. Ax = -1 - (-3) = 2, Ay = -2 - 2 = -4; d = ^(Ax) 2 + (Ay) 2 = ^4+ 16 = 2 a/5 2. Ax = -3 - (-1) = -2, Ay = 2 - (-2) = 4; d = ^/(-2) 2 + 4 2 = 2 a/5 3. Ax = -8.1 - (-3.2) = -4.9, Ay = -2 - (-2) = 0; d = ^/(-4.9) 2 + 2 = 4.S 4. Ax = 0- \pl= -s/l, Ay = 1.5-4 = -2.5; d 5. Circle with center (0, 0) and radius 1. 2) + (-2.5) 2 = a/^25 6. Circle with center (0, 0) and radius y 2. 7. Disk (i.e., circle together with its interior points) with center (0, 0) and radius y 3. 8. The origin (a single point). 9. m Ay -1-2 Ax -2-(-l) 10. m Ay -2-1 Ax 2 -(-2) perpendicular slope perpendicular slope = | y 4 3 2 AN 1 -4 -2 -1 -2 -3 -4 2 4 Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle Chapter 1 Preliminaries Ay _ 3-3 _ q perpendicular slope does not exist «•»=£- -1-2 12 - m =S = 3y^;no slope perpendicular slope = fi(-l, 3) A(2, 3) * 3 2 >' = 3 Slope = 1 1 1 *x 4 3 2 B 1 -4 - 2 -1 2 4 A - 2 -3 -4 13. (a) x = -1 (b) y = | 14. (a) x = y/2 (b) y = 1.3 15. (a) x = (b) y=-y/2 16. (a) x = — it (b) y = 17. P(-l,l),m=-l => y-1 = -l(x-(-l)) => y = -x = ix-4 18. P(2, -3), m = i => y - (-3) = \ (x - 2) => y = 5 x 19. P(3,4),Q(-2,5) => m=^ = ^3 = -i => y_4 = -I(x-3) 20. P(-8,0),Q(-1,3) => m = £* = ^^f^ = § => y- = § (x- (-8)) => y= fx+f ' = -** I v J- 23 21. m = -|,b = 6 => y = -| 23. m = 0,P(-12,-9) => y = . + 6 22. m=i,b=-3 =4> y=i x -3 24. Noslope, P(|,4) =>• x= 1 > y = 4x + 4 i the line => m = ^ = -&=4 = Ax —1—0 =>• y = -f x+ 1 25. a=-l,b = 4 => (0,4) and (-1,0) are on i a = 2, b = -6 =>• (2, 0) and (0, -6) are on the line => m = ^ = z£^ =3 ^ y = 3x _ 6 27. P(5,-l), L: 2x + 5y = 15 => m L = -| => parallel line is y - (-1) = -| (x - 5) => y = v / 3^m L = -^^ parallel line is y - 2 = -^ ( x _ (-1/2)) => y = P(4, 10), L: 6x - 3y = 5 => m L = 2 =4> m ± = -5 =>• perpendicular line is y - 10 = — | (x — 4) => y 30. P(0, 1), L: 8x - 13y = 13 =4> m L = ^ => m ± = -^ =4> perpendicular line is y = -^ x + 1 26 28. P (-a/2,2) ,L: A/2x + 5y 29. -^x + ^ 5 S+ 5 4x4- 12 Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 31. x-intercept = 4, y-intercept = 3 Section 1.2 Lines, Circles and Parabolas 32. x-intercept = —4, y-intercept = —2 t i V. 3*+4;y=12 2 1 1 2 3 is. 33. x-intercept = y\3, y-intercept = — y 2 t * . l/ 2 -1 "/ ~E.x-~&y = -fc -2 34. x-intercept = —2, y-intercept = 3 4 / 2 / 1 1.5x-y. -3 -4 , -1 -2 -3 -4 2 4 35. Ax + By = Ci 4* y = -§ x + §■ and Bx - Ay = C 2 O y= |x- &. Since (-§) (|) = -lis the product of the slopes, the lines are perpendicular. 36. Ax + By = Ci <4> y = — ^ x + ^ and Ax + By = C 2 O y = — | x + ^. Since the lines have the same slope — § , they are parallel. 37. New position = (x old + Ax, y ok , + Ay) = (-2 + 5,3 + (-6)) = (3, -3). 38. New position = (x old + Ax, y„ kl + Ay) = (6 + (-6), + 0) = (0, 0). 39. Ax = 5, Ay = 6, B(3, -3). Let A = (x, y). Then Ax = x 2 - x : => 5 = 3 - x => x = -2 and Ay = y 2 - yi => 6 = -3 - y => y = -9. Therefore, A = (-2, -9). 40. Ax = 1 - 1 = 0, Ay = - = Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 8 Chapter 1 Preliminaries 41. C(0, 2), a = 2 => x 2 + (y - 2) 2 = 4 . J0.4) [ C(0,2)i h J (0,0)/ -2 -1 1 ' 1 2 * 42. C(-3,0),a = 3 => (x + 3) 2 + y 2 43. C(-l,5),a= V10 => (x + l) 2 + (y - 5) 2 = 10 44. C(l, 1), a = 1/2 =>• (x - l) 2 + (y - l) 2 = 2 x = => (0 - l) 2 + (y - l) 2 = 2 =>■ (y - l) 2 = 1 => y — 1 = ±1 => y = or y = 2. Similarly, y = =4> x = 0orx = 2 45. c(- v / 3,-2),a = 2 => (x + ^3) + (y + 2) 2 = 4, x = => (0 + a/3~) + (y + 2) 2 = 4 => (y + 2) 2 = 1 => y + 2= ±1 =4> y = - 1 or y = -3. Also, y = => U + a/3) + (0 + 2) 2 = 4 => (x + 1/3) = => x= -a/3 46. C (3, |), a = 5 =>• (x - 3) 2 + (y - i) 2 = 25, so x = =>• (0-3) 2 + (y- |) 2 = 25 => (y-i) 2 = 16 => y-|= ±4 => y=| ory = -|. Also, y = => (x - 3) 2 + (0 - \f = 25 => (x - 3) 2 = f => x - 3 = ± *-& => x = 3±^- Copyright (c) 2006 Pearson Education, Inc (0,-3) Section 1.2 Lines, Circles and Parabolas 47. x 2 + y 2 + 4x - 4y + 4 = =>■ x 2 + Ax + y 2 - 4y = -4 =>■ x 2 + 4x + 4 + y 2 - 4y + 4 = 4 => (x + 2) 2 + (y - 2) 2 = 4 => C = (-2, 2), a frt + 2) 2 + G> y -2) 2 = 4 -3 • (0,2) -2 C(-2, 2) (-2, o) y -4 -i -2 -1 48. x 2 + y 2 - 8x + 4y + 16 = =4> x 2 - 8x + y 2 + 4y = - 16 =>• x 2 - 8x + 16 + y 2 + 4y + 4 = = 4 => (x - 4) 2 + (y + 2) 2 = 4 =>• C = (4, -2), a = 2. 49. x 2 + y 2 - 3y - 4 = =s> x 2 + y 2 - 3y => x 2 + y 2 - 3y + § = f v2 + (y-|)' 25 4 c=(o,|) a 50. x 2 + y 2 -4x-| = = =>■ x 2 - - 4x + y 2 = 9 4 =>■ x 2 - - 4x + 4 + y 2 = 25 4 =^ (x- - 2) 2 + y 2 -- 25 4 =*> C = = (2,0),a = 5 2 - (4.5.0) 51. x 2 + y 2 -4x + 4y = =>■ x 2 - 4x + y 2 + 4y = =4> x 2 - 4x + 4 + y 2 + 4y + 4 = 8 => (x - 2) 2 + (y + 2) 2 = 8 => C(2, -2), a = s/%. , * (x-2f + (y + 2f = $ (0,0) J^ \ (4,0) -1 / l-\ 1 2 3 i\ 5 -2 V-3 * C(2,-2) / -41 (-4, 0) / Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle 10 Chapter 1 Preliminaries 52. x 2 + y 2 + 2x = 3 => x 2 + 2x + 1 - y 2 =4 (x+l) 2 + y 2 = 4 C = (-1,0), a = 2. 53. x = -f 2a -2 2(1) 1 => y = (l) 2 - 2(1) - 3 = -4 =>. V = (l,-4). Ifx = 0theny = -3. Also, y = => x 2 -2x-3 = =>■ (x - 3)(x + 1) = =>■ x = 3or x = — 1. Axis of parabola is x — 1 . V(l,-4) 54. x = - £ 4 2(1) => y = (-2) 2 +4(-2) + 3 = -l ^ V = (-2,-l). Ifx = 0theny = 3. Also, y = => x 2 +4x + 3 = => (x + l)(x + 3) = => x = -1 or x = —3. Axis of parabola is x = —2. 55 x - _ A - 4_ - ? JJ - A — 2a — 2(-l) ~~ z - => y = -(2) 2 + 4(2) = 4 =>■ V = (2,4). Ifx = 0theny = 0. Also, y = =^ -x 2 + 4x = =>■ -x(x - 4) = => x = 4orx = 0. Axis of parabola is x = 2. V(2, 4) 56. * 2a 2(-l) =► y = -(2) 2 +4(2)-5 = -l =>■ v = (2,-1). Ifx = 0theny = Also, y = = => -x 2 + 4x - 5 = =► x 2 - 4x + 5 = =>• x = ^f 4 no x intercepts. Axis of parabola is x = 2. Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Section 1.2 Lines, Circles and Parabolas 11 57. x = - £ -3 -6 2(-l) (-3) 2 -6(-3)-5 = 4 =>■ V = (-3,4). Ifx = Otheny = -5. Also, y = => -x 2 - 6x - 5 = => (x + 5)(x + 1) = =4> x = -5 or x = — 1. Axis of parabola is x = —3. 1(0,-5) 58. x=- 2a -1 2(2) 2Q) 2 -i + 3=f => y =>. V= (i,f) . Ifx = 0theny = 3. Also, y = => 2x 2 -x + 3 = l±V-23 no x intercepts. Axis of parabola is x 59. x 1 2a 2(1/2) => y = |(-i) 2 + (-i) + 4 = | =>■ V= (-1,|) ■ Ifx = 0theny = 4. Also, y = => |x 2 + x + 4 = =4> x = z — j^— =4- nox intercepts. Axis of parabola is x = — 1 . 60. x = - £■ 2a 2(-l/4) _ ^ => y = - 1 (4) 2 + 2(4) + 4 = 8 =>. V = (4, 8) . If x = then y = 4. Also, y = => -jX 2 + 2x + 4 = => *=^# =4 + 4^2. Axis of parabola is x = 4. (-2, 4) ** V(-l,7/2) (0,4) ;y=^* 2 + j: + 4 5> (4 + 4 >/2.o) 61. The points that lie outside the circle with center (0, 0) and radius y 7. 62. The points that lie inside the circle with center (0, 0) and radius y 5. 63. The points that lie on or inside the circle with center (1, 0) and radius 2. 64. The points lying on or outside the circle with center (0, 2) and radius 2. 65. The points lying outside the circle with center (0, 0) and radius 1, but inside the circle with center (0, 0), and radius 2 (i.e., a washer). Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 12 Chapter 1 Preliminaries 66. The points on or inside the circle centered at (0, 0) with radius 2 and on or inside the circle centered at (—2, 0) with radius 2. 67. x 2 + y 2 + 6y < => x 2 + (y + 3) 2 < 9. The interior points of the circle centered at (0, —3) with radius 3, but above the line y = -3. 68. x 2 + y 2 - 4x + 2y > 4 => (x - 2) 2 + (y + l) 2 > The points exterior to the circle centered at (2,-1) with radius 3 and to the right of the line x = 2. 69. (x + 2) 2 + (y - l) 2 < 6 71. x 2 + y 2 < 2, x > 1 73. x 2 + y 2 = 1 and y = 2x => 1 = x 2 + 4x 2 = 5x 2 =* (x = -*- andy = -^) or (x = --J- and y Thus ' A (7i'7;)' B (-7;'-7i) arethe points of intersection. 70. (x + 4) 2 + (y - 2) 2 > 16 72. x 2 + y 2 > 4, (x - l) 2 + (y - 3) 2 < 10 *) 2 2 Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Section 1.2 Lines, Circles, and Parabolas 13 74. x + y = 1 and (x - l) 2 + y 2 = => 1 = (-y) 2 + y 2 = 2y 2 4= and x = 1 ! y "75 """ A -^75. (y = -^ andx=l + 4^). Thus, A ( 1 -75'73) andB ( 1 + 73'-7i) are intersection points. ( x -1) 2 H.y 2 = l 75. y — x = 1 and y = x =>• x — x = 1 => x 2 -x-l=0 => x = ^ . Ifx 1 + %/s then y = x + 1 If x = ±^ , then y = x + 1 3+y/5 2 3-n/5 Thus ,A( 1+V5 3+^ andB^-VS 3-V5 2 > 2 are the intersection points 76. y = — x and y = — (x — l) 2 =>- (x — l) 2 =>■ x 2 - 3x + 1 x = ^ . If 2 3+A/5 then y then y /5-3 If 3+V5 Thus,A(^ J 4-)andB(^,-^) 2 ' 2 are the intersection points. 77. y = 2x 2 - 1 = -x 2 => 3x => x 4- and y = — i or x = \- and y = — I Vi J 3 V3 J 3 Thus, A (4- , - 1 J and B (- 4; , - i J are the intersection points. Copffigl (c| 1 Pearson Educate he, publishing as Pearson Addison-Wesle 14 Chapter 1 Preliminaries 78. y (x - l) 2 3x 2 -2x+ 1 = 3x 2 4 = (3x - 2)(x - 2) 1 , or x = | and x = 2 and y = \ % = \. Thus, A(2,l) and B(§,± are the intersection points. 79. x 2 + y 2 = 1 = (x - l) 2 - =>. X 2 = (x- l) 2 =x 2 - => = -2x + 1 => x y 2x+ 1 = 1 Hence 1 | ory ± /3 Thus, A(±,^) mxdBU,-^) are the intersection points. A (x-l) 2 +y 2 -l 80. x 2 -J- " 2 x 2 + y => y 2 y y = or y -y 2 i. r = i => y(y - 1) = If y = 1, then x 2 = 1 — y 2 = or x = 0. If y = 0, then x 2 = 1 — y 2 = 1 or x = ± 1. Thus, A(0, 1), B(l, 0), and C(-l, 0) are the intersection points. 1 -x' $1. (a) A«(69°,0in), B w (68°, .4 in) (b) A w (68°, .4 in), B « (10°, 4 in) (c) A w (10°, 4 in), B w (5°, 4.6 in) 68° - 69° m .4 -0 10° -68° 4- -.4 5°- 10° 4.6-4 -2.5°/in. -16.17in. ^8.3°/in. 82. The time rate of heat transfer across a material, -^ , is directly proportional to the cross-sectional area, A, of the material, to the temperature gradient across the material, ^ (the slopes from the previous problem), and to a constant characteristic of the material. 4j = -kA^J =4- k = — Hj- . Note that 4j and ^ are of opposite sign because heat flow is toward lower temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are not changing), we may define another constant, K, characteristics of the material: K = — -^ . Using the values of ^ from At the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the poorest insulator, with K = 0.4. 83. p = kd + 1 and p = 10.94 at d = 100 pressure equation so that d = 50 t _ 10.94-1 K 100 p = (0.0994)(50) 0.0994. Then p = 0.0994d + 1 is the diver's 1 = 5.97 atmospheres. 84. The line of incidence passes through (0, 1) and (1,0) => The line of reflection passes through (1, 0) and (2, 1) =$■ m = ^5j = 1 =** y — = l(x — 1) =^> y = x— lis the line of reflection. Copyright (c) 2006 Pearson Education 160 160 9 Section 1.2 Lines, Circles, and Parabolas 15 or F = —40° gives the same numerical reading. 86. m 37.1 _14 Ax Ax = ~j. Therefore, distance between first and last rows is a/(14) 2 + (4lr) « 40.25 ft. 87. length AB = a/(5 - l) 2 + (5 - 2) 2 = a/T6 length AC = a/(4 - l) 2 + (-2 - 2) 2 length BC = a/(4 - 5) 2 + (-2 - 5) 2 9 = 5 a/9 +16 = 5 V / TT49= a/50 5V2^5 88. length AB = J(l - 0) 2 + (a/3 - o) = a/1+3 = 2 length AC = a/(2 - 0) 2 + (0 - 0) 2 = a/4 + = 2 length BC \y(2-D 2 +(0-A/3) 2 = A/ 1+3 a/i 2 + 4 2 = a/17 and length BC = a/(Ax) 2 + (Ay) 2 = a/4 2 + l 2 89. Length AB = a/(Ax) 2 + (Ay) 2 = a/ I 2 + 4 2 = a/ 17 and length BC = a/(Ax) 2 + (Ay) 2 = a/4 2 + l 2 = a/ 17. Also, slope AB = ^y and slope BC = 4, so AB ± BC. Thus, the points are vertices of a square. The coordinate increments from the fourth vertex D(x, y) to A must equal the increments from CtoB => 2 — x = Ax = 4 and — 1 — y = Ay = 1 =>■ x = —2 and y = —2. Thus D(— 2, —2) is the fourth vertex. 90. Let A = (x, 2) and C = (9, y) => B = (x, y). Then 9 - x = |AD| and 2 - y = |DC| => 2(9 - x) + 2(2 - y) = 56 and 9 - x = 3(2 - y) =* 2(3(2 - y)) + 2(2 - y) = 56 => y = -5 => 9 - x = 3(2 - (-5)) =4> x = -12. Therefore, A = (-12, 2), C = (9, -5), and B = (-12, -5). 91. Let A(-l, 1), B(2, 3), and C(2, 0) denote the points. Since BC is vertical and has length |BC| = 3, let Di(— 1,4) be located vertically upward from A and D2(— 1, —2) be located vertically downward from A so that |BC| = |ADi| = |AD 2 | = 3. Denote the point Da(x, y). Since the slope of AB equals the slope of CD 3 we have £E§ = ~ 3 => 3y-9 = -x + 2or x + 3y = 11. Likewise, the slope of AC equals the slope of BD 3 so that |f| = § => 3y = 2x - 4 or 2x - 3y = 4. x + 3y Solving the system of equations 2x-3y "} (-1,-2)* -2 • (5,2) we find x = 5 and y = 2 yielding the vertex D3(5, 2). 92. Let (x,y),x^0 and/or y 7^ be a point on the coordinate plane. The slope, m, of the segment (0, 0) to (x, y) is - . A 90° rotation gives a segment with slope m' = — j- = — | . If this segment has length equal to the original segment, its endpoint will be (— y, x) or (y, — x), the first of these corresponds to a counter-clockwise rotation, the latter to a clockwise rotation. (a) (-1,4); (b) (3,-2); (c) (5,2); (d) (0,x); Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 16 Chapter 1 Preliminaries (e) (-y,0); (f) (-y,x); (g) (3,-10) 93. 2x + ky = 3 has slope — | and 4x + y = 1 has slope —4. The lines are perpendicular when — | (—4) 1 or k = — 8 and parallel when -4 or k 2' 94. At the point of intersection, 2x + 4y = 6 and 2x — 3y = — 1 . Subtracting these equations we find 7y = 7 or y = 1. Substitution into either equation gives x = 1 =$■ (1,1) is the intersection point. The line through (1, 1) and (1, 2) is vertical with equation x = 1. 95. Let M(a, b) be the midpoint. Since the two triangles shown in the figure are congruent, the value a must lie midway between Xi and X2, so a *l+*2 2 ■ Similarly, b= 2d£>. ....P^.y,) M(a,b) Q(x 2 ,y 2 ) 96. (a) L has slope 1 so M is the line through P(2, 1) with slope —1; or the line y = point, Q, we have equal y-values, y = x + 2 = — x + 3. Thus, 2x = 1 or x 'I 1) ,2> 2) The distance from P to L = the distance from P to Q x + 3. At the intersection I . Hence Q has coordinates . V2 i)' (b) L has slope — | so M has slope | and M has the equation 4y — 3x = 12. We can rewrite the equations of the lines as L: x + | y = 3 and M into either equation gives x -x+ fy 4. Adding these we get y| y = 7 so y || . Substitution 4 (S4\ 3 I 25/ 4 = || so that Q (si' If) * s me P omt of intersection. The distance fromPtoL=^(4-i|) 2 + (6-l) 2 = f. (c) M is a horizontal line with equation y = b. The intersection point of L and M is Q(— 1, b). Thus, the distance from P to L is ^/(a + l) 2 + 2 = |a + 1 1 . (d) If B = and A^O, then the distance from P to L is | j - x 1 as in (c). Similarly, if A = and B^O, the A distance is I £ I A yo | . If both A and B are ^ then L has slope -gSoM has slope |. Thus, L: Ax + By = C and M: — Bx + Ay = — Bxo + Ayo. Solving these equations simultaneously we find the point of intersection Q(x, y) with x AC-B (Ayp-Bxp) A 2 +B 2 and y BC+A(Ay -Bx ) A 2 +B 2 The distance from P to Q equals ^(Ax) 2 + (Ay) 2 , where (Ax) 2 = ^ (A 2 +B 2 )-AC+AB yo -B% y A 2 (Ax +By +C) : (A 2 +B 2 ) 2 , and (Ay) 2 = ( yo (A 2 +B 2 )-BC-A 2 yo+ABx ' A 2 +B 2 B 2 (Axp+Byo+C) 2 (A 2 +B 2 ) 2 Thus, v/(Ax) 2 + (Ay) 2 (Axp+Byp+C) 2 A 2 +B 2 |Ax +Byo+C| v'a 2 +b 2 1.3 FUNCTIONS AND THEIR GRAPHS 1. domain = (—oo, oo); range = [1, oo) 2. domain = [0, oo); range = (— oo, 1] 3. domain = (0, oo); y in range => y=^ r ,t>0 =^ y 2 = \ and y > =^ y can be any positive real number =^ range = (0, oo). Copyright (c) 2006 Pearson Education 4. domain = [0, oo); y in range and smaller positive real number =>■ range = (0, 1] Section 1.3 Functions and Their Graphs 17 y = — A- , t > 0. If t = 0, then y = 1 and as t increases, y becomes a smaller J i+\A 5. 4 — z 2 = (2 — z)(2 + z) > <^ z 6 [—2, 2] = domain. Largest value is g(0) = y 4 = 2 and smallest value is g(-2) = g(2) = v/0 = =► range = [0, 2]. 6. domain = (—2, 2) from Exercise 5; smallest value is g(0) = \ and as < z increases to 2, g(z) gets larger and larger (also true as z < decreases to —2) =4> range = [|, oo) . 7. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. 8. (a) Not the graph of a function of x since it fails the vertical line test, (b) Not the graph of a function of x since it fails the vertical line test. 9. y 1 => - - 1 > =>• x < 1 and x > 0. So, (a) No (x > 0); (b) No; division by undefined; (c) No; if x > 1, i < 1 => i - 1 < 0; (d) (0, 1] 10. y = a/2 - a/x => 2 - a/x > => a/x > and a/x < 2. a/x > => x > and a/x < 2 =4> x < 4. So, < x < 4. (a) No; (b) No; (c) [0, 4] 11. base = x; (height) 2 + (f) = x 2 => height = \- x; area is a perimeter is p(x) = x + x + x = 3x. (x) = 1 (base)(height) = \ (x) (^x) = ^ x 2 ; 12. s = side length => s 2 + s 2 = d 2 =^ s = A- ; and area is a = s 2 =^> a = | d 2 13. Let D = diagonal of a face of the cube and t = the length of an edge. Then I + D = d" and (by Exercise 10) D 2 = 2l 2 => 3f = d 2 ^ £ = A ■ The surface area is 6f = ^# = 2d 2 and the volume is£ 3 =(f) 3/2 d 3 14. The coordinates of P are (x, ^x) so the slope of the line joining P to the origin is m = ^ = A- (x > 0). Thus, 16. The domain is (— oo, oo). f( x) 3 f(x) = l-2x-x 2 X~ "N, 2 / - 1 -3 / -2 -1 \ 1 -1 -2 Copffigl (c) 1 Pearson Etation, Ire, publishing as Pearson Addison-Wesle 18 Chapter 1 Preliminaries 17. The domain is ( — oo, oo). 19. The domain is ( — oo, 0) U (0, oo). F(l) = 12 3 4 18. The domain is (— oo, 0]. 20. The domain is (— oo, 0) U (0, oo). f(x) 1 21. Neither graph passes the vertical line test (a) 22. Neither graph passes the vertical line test (a) (b) (b) f x + y = i | [y=i-x| < or > <^> < or > x + y |x + y| = 1 <^> I or [ x + y =_lj ^y= l xj Copffigl (c) 1 Pearson M% } k, publishing as Pearson Addison-Wesle Section 1.3 Functions and Their Graphs 19 23. X 1 2 y 1 24. X i 2 y i r 1 -x , o<x<" l 2-x,1 <x<2 25. y 3 - x, x < 1 2x, 1 < x 2 1 . F(x) = 3 - x, x<l 2x, x>l -3 -2 -1 1 2 3 • (a) Line through (0, 0) and (1, 1): y = x Line through (1, 1) and (2. 0): y = -x + 2 fl , , f x, < x < 1 W -\-x + 2, 1 < x < 2 '2, < x < 1 (b) f(x) = < 0, 1 < x < 2 2, 2 < x < 3 0, 3 < x < 4 26. y \> x<0 x, < x 28. (a) Line through (0, 2) and (2, 0): y Line through (2, 1) and (5, 0): m J -x 4- 2, < x < 2 f(x)= \ -ix+jj, 2<x<5 (b) Line through (-1, 0) and (0, -3): m Line through (0, 3) and (2, —1): m = ' -3x-3, -1 < x < o- 1 5-2 3 -3-0 " 0-(-l) -1-3 2-0 f(x) -2x- < x < 2 so y |(x-2) + l -3, so y = — 3x — 3 = -2, soy = -2x4-3 29. (a) Line through (-1, 1) and (0, 0): y = -x Line through (0, 1) and (1, 1): y = 1 Line through (1, 1) and (3, 0): m = §=i 2 so y •i(x- 1)4-1 —x 1 f(x) = I 2^ ' 2 (b) Line through (—2, -1 < x < < x < 1 1< x < 3 -l)and(0, 0):y oX Line through (0, 2) and (1, 0): y = -2x 4- 2 Line through (1, —1) and (3, —1): y = — 1 - „x + „ Copyright (c) 2006 Pearson Education 20 Chapter 1 Preliminaries f(x) Iv 2 x -2 < x < -2x + 2 < x < 1 ^ -1 1 < x < 3 30. (a) Line through (f , 0) and (T, 1): m 1-0 T-(T/2) so y f(x) (b) f(x) = i 0, < x < | ix-1, I < x < T A, < x < | -A, < x < T 3T A, T < x < -A, f < x < 2T • = ix-i 31. (a) From the graph, | > 1 xe (-2,0)U(4,oo) (b) | > 1 => 5-1-4 >Q 2 x>0: | - 1 - 1 > ^ ^f=^ > =^> x > 4 since x is positive; 4 >Q => ^!4^8 <0 x < 0: f - 1 2x x < —2 since x is negative; sign of (x - 4)(x + 2) ^t— -2 4 Solution interval: (-2, 0) U (4, oo) (x-4)(x+2) n 2x ^ U (x-4)(x+2) n 2x v u g{x) = \*1 f(x) -\ 32. (a) From the graph, ^y < ^y =4> x e (-oo, -5) U (-1, 1) (b) Casex<-1: £ < £ => ^ > 2 =>• 3x + 3 < 2x - 2 => x < -5. Thus, x G (— oo, —5) solves the inequality. Case - 1 < x < 1: r^r < 2 3(i+l) <2 x-l x+1 x-1 =4> 3x + 3 > 2x — 2 =4> x > — 5 which is true if x > — 1. Thus, x G (—1,1) solves the inequality. Case l<x: ^t < ^y => 3x + 3<2x-2 => x<-5 which is never true if 1 < x, so no solution here. In conclusion, x e (— oo, —5) U (— 1, 1). 33. (a) [xj =0forxe [0,1) (b) |Y| =0forxe (-1,0] 34. L X J = |~ x l only when x is an integer. 35. For any real number x, n < x < n + 1, where n is an integer. Now: n<x<n+l=>— (n + 1) < — x < — n. By definition: |~— x] = — n and |_xj = n => — |_xj = — n. So |~— x] = — |_xj for all x e 5R. Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 1.3 Functions and Their Graphs 21 36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part. -3- -i -2| f(x}. -3 |xj, x>.0 r*T > x < ° 37. v = f(x) = x(14 - 2x)(22 - 2x) = 4x 3 - 72x 2 + 308x; < x < 7. 38. (a) Let h = height of the triangle. Since the triangle is isosceles, AB 2 + AB ' AB '2. So, h 2 + l 2 = (y/2) => h = 1 => B is at (0, 1) =>■ slope of AB = -1 => The equation of AB is y = f(x) = -i+l;xe [0, 1]. (b) A(x) = 2xy = 2x(-x + 1) = -2x 2 + 2x; x E [0, 1]. 39. (a) Because the circumference of the original circle was 8tt and a piece of length x was removed. X 2rr (b)r=^=4-i (c) h = v/16^7 2 " = 7l6-(4-^) 2 = ^16- (16- £+*,)-. A* - ^i - - /il- - --- ^^ (d) V=^r 2 h=l7r( 2 4ir 2 V 47r 2 47r 2 i ,,2 1, : „( 8ir-x \ 2 \/l67rx - x 2 _ (87r - x) 2 Vl67rx - x 2 2?r — 24?r 2 40. (a) Note that 2 mi = 10,560 ft, so there are V800 2 + x 2 feet of river cable at $180 per foot and (10, 560 - x) feet of land cable at $100 per foot. The cost is C(x) = 180 v / 800 2 ~+x 2 + 100(10, 560 - x). (b) C(0) = $1,200,000 C(500) « $1,175, 812 C(1000) « $1,186, 512 C(1500) « $1,212,000 C(2000) w $1,243, 732 C(2500) « $1,278,479 C(3000) w $1,314,870 Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P. 41. A curve symmetric about the x-axis will not pass the vertical line test because the points (x, y) and (x, — y) lie on the same vertical line. The graph of the function y = f(x) = is the x-axis, a horizontal line for which there is a single y-value, 0, for any x. 42. Pick 1 1, for example: 11 + 5 = 16 -» 2 • 16 = 32 ->■ 32 - 6 = 26 f(x) 2(x+5)-6 2 = x, the number you started with. Y = 13— >13 — 2 = 11, the original number. Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 22 Chapter 1 Preliminaries 1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS 1. (a) linear, polynomial of degree 1, algebraic. (c) rational, algebraic. (b) power, algebraic, (d) exponential. 2. (a) polynomial of degree 4, algebraic, (c) algebraic. (b) exponential. (d) power, algebraic. 3. (a) rational, algebraic, (c) trigonometric. (b) algebraic, (d) logarithmic. 4. (a) logarithmic. (c) exponential. (b) algebraic, (d) trigonometric. 5. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 6. (a) Graph f because it is linear. (b) Graph g because it contains (0, 1). (c) Graph h because it is a nonlinear odd function. 7. Symmetric about the origin Dec: — oo < x < oo Inc: nowhere Symmetric about the y-axis Dec: — oo < x < Inc: < x < oo 9. Symmetric about the origin Dec: nowhere Inc: — oo < x < < x < oo 10. Symmetric about the y-axis Dec: < x < oo Inc: — oo < x < -2-1 -2 -li Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle Section 1.4 Identifying Functions; Mathematical Models 23 11. Symmetric about the y-axis Dec: — oo < x < Inc: < x < oo 12. No symmetry Dec: — oo < x < Inc: nowhere ■vn 13. Symmetric about the origin Dec: nowhere Inc: — oo < x < oo 14. No symmetry Dec: < x < oo Inc: nowhere t . 1 ~y- i 1/8 _ 2 / ^■""-1/8 / -1 ■ 15. No symmetry Dec: < x < oo Inc: nowhere y . , \ 2 3 -l -2 \ y ^ 3/2 -X -3 -4 -5 16. No symmetry Dec: — oo < x < Inc: nowhere y 25 20 15 3/2 X. y-H> N. 10 5 -8 -6 -4 -2 -5 -10 Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 24 Chapter 1 Preliminaries 17. Symmetric about the y-axis Dec: — oo < x < Inc: < x < oo 18. Symmetric about the y-axis Dec: < x < oo Inc: — oo < x < y 10 7.5 s 2.5 -20 -io ; -2/5 s " 5 10 20 -7.5 t~ 2/3 -X -10 19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even. 20. f(x) = x- 5 = \ and f(-x) = (-x) -5 _ 1 (-x) 5 '^\ = — f(x). Thus the function is odd. 21. Since f(x) = x 2 + 1 = (— x) + 1 = — f(x). The function is even. 22. Since [f(x) = x 2 + x] ^ [f(— x) = (— x) — x] and [f(x) = x 2 + x] 7^ [— f(x) = — (x) — x] the function is neither even nor odd. 23. Since g(x) = x 3 + x, g(— x) = — x 3 — x = — (x 3 + x) = — g(x). So the function is odd. 24. g(x) = x 4 + 3x 2 + 1 = (— x) + 3(— x) — 1 = g(— x), thus the function is even. 25. g(x) = ^rzrj = ,_ I2, = g( — x )- Thus the function is even. 26. g(x) = ^ri; g(-x) = - j^ = g(-x). So the function is odd. 27. h(t) = ^j; h(-t) = ~^zrj; -h(t) = j-L. Since h(t) ^ -h(t) and h(t) ^ h(-t), the function is neither even nor odd. 28. Since 1 1 3 | = | (-t) 3 |, h(t) = h(-t) and the function is even. 29. h(t) = 2t + 1, h(-t) = -2t + 1. So h(t) ^ h(-t). -h(t) = -2t - 1, so h(t) ^ -h(t). The function is neither even nor odd. 30. h(t) = 2| 1 1 + 1 and h(-t) = 2| -t | + 1 = 2| 1 1 + 1. So h(t) = h(-t) and the function is even. 31. (a) y = 0.166.t The graph supports the assumption that y is proportional to x. The constant of proportionality is estimated from the slope of the regression line, which is 0.166. Section 1.4 Identifying Functions; Mathematical Models 25 (b) The graph supports the assumption that y is proportional to x 1 / 2 . The constant of proportionality is estimated from the slope of the regression line, which is 2.03. 32. (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the regression line. y 500 1000 1500 2000 2500" The graphs support the assumption that y is proportional to 3 X . The constant of proportionality is estimated from the slope of the regression line, which is 5.00. (b) The graph supports the assumption that y is proportional to In x. The constant of proportionality is extimated from the slope of the regression line, which is 2.99. 20 15 10 5 In x 33. (a) The scatterplot of y = reaction distance versus x = speed is 10 20 30 40 50 60 70 80 90100 Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approximately 1.1. Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle 26 Chapter 1 Preliminaries (b) Calculate x' = speed squared. The scatterplot of x'versus y = braking distance is: 400 6400 Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approximately 0.059. 34. Kepler's 3rd Law is T(days) = 0.41R 3 ' 2 , R in millions of miles. "Quaoar" is 4 x 10 9 miles from Earth, or about 4 x 10 9 + 93 x 10 6 « 4 x 10 9 miles from the sun. Let R = 4000 (millions of miles) and T = (0.41)(4000) 3/2 days w 103, 723 days. 35. (a) y ■ 10 9 • 8 • 7 • 6 • 5 • 4 3 • m • 2 1 • • 1 2 1 4 5 6 7 8 9 10 8.741 -0 The hypothesis is reasonable. (b) The constant of proportionality is the slope of the line ~ (c) y(in.) = (0. 87 in./unit mass)(13 unit mass) = 11.31 in. in./unit mass = 0.874 in./unit mass. 36. (a) (b) 300 200 100 300 200 100 • • 500 1000 1500 20000 40000 60000 Graph (b) suggests that y = kx 3 is the better model. This graph is more linear than is graph (a). 1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 1. D, : -oo < x < oo, D g : x > 1 => D r+g = D lg : x > 1. R, : -oo < y < oo, R g : y > 0, R 1+g : y > 1, R lg : y > 2. D, : x + 1 > => x > -1, D g : x - 1 > => x > 1. Therefore D f+g = D fg : x > 1. Rf = R g : y > 0, R f+g : y > y/l, R lg : y > Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Section 1.5 Combining Functions; Shifting and Scaling Graphs 27 3. D f : — oo < x < oo, D g : — oo < x < oo =>• D,y g : — oo < x < oo since g(x) ^ for any x; D g/ , : — oo < x < oo since f(x) / for any x. R, : y = 2, R„: y > 1, R f/g : < y < 2, R g/f : y > \ 4. D f : -oo < x < oo, D g : x > =4> D f/g : x > since g(x) ^ for any x > 0; D g/f : x > since f(x) ^ for any x > 0. R f : y = 1, R E : y > 1, R l/g : < y < 1, R g/f : y > 1 5. (a (b (c (d (c (f) ( (h 6. (a (b (c (d (c (f) ( (h 7. (a (b (c (d (e (f) 8. (a (b (c (d (c (f) 9. (a (c (c 10. (a (c (c f(g(0)) = f(-3) = 2 g(f(0)) = g(5) = 22 f(g(x)) = f(x 2 - 3) = x 2 - 3 + 5 = x 2 + 2 g(f(x)) = g(x + 5) = (x + 5) 2 - 3 = x 2 + lOx + 22 f(f(-5)) = f(0) = 5 g(g(2)) = g(l) = -2 f(f(x)) = f(x + 5) = (x + 5) + 5 = x + 10 g(g(x)) = g(x 2 - 3) = (x 2 - 3) 2 - 3 = x 4 - 6x 2 + 6 f(g(i» g(f(D) '(f) {-\) ffe(x))=f(ji T ): g(f(x)) = g(x - 1) f(f(2)) = f(l) = g(g(2)) l x+I x+1 1 (x-l)+l 8 (J) f(f(x)) = f(x - 1) = (X - 1) - 1 g(g(x)) = g (rrr) = -+1 x+1 x + 2 (x/ -1 andx^ -2) 5- ± u(v(f(x))) = u(v(I))=u(i)=4(i) u(f(v(x))) = u (f (x 2 )) = u (i) = 4 (i) - 5 v(u(f(x))) = v(u(I))=v(4(i)-5) = ( 4 -5) v(f(u(x))) = v(f(4x - 5)) = v (3^5) f(u(v(x))) = f(u(x 2 ))=f(4(x 2 )-5 f(v(u(x))) = f(v(4x - 5)) = f ((4x - 5) : .x lix^li 1 4x 2 -5 \2\ _ 1 2 (4x - 5) 2 h(g(f(x))) = h (g (v^)) = h (^) = 4 (^) h(f(g(x))) = h (f (I)) = h (v% = VJ - 8 g(h(f(x))) = g (h (y^)) = g (4^/x" - 8) g(f(h(x))) = g(f(4x - 8)) = g (^4x-8) f(g(h(x))) = f(g(4x - 8)) = f (^) = f(x - 2) f(h(g(x))) = f(h(|))=f(4(|)-8) y = f(g(x)) y = g(g(x)) y = g(h(f(x))) y = fQ(x)) y = h(h(x)) y=j(g(f(x))) 8 = 2^ ygx - X 4 'x-2 y/x-2 2 f(x - 8) = Vx" (b) y=j(g(x)) (d) y = jQ(x)) (f) y = hG(f(x))) (b) y = h(g(x)) = g(h(x)) (d) y = f(f(x)) (f) y = g(f(h(x))) Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 28 Chapter 1 Preliminaries 11. g(x) f(x) (f°g)(x) (a) x - 7 (b) x + 2 (c) x 2 (d) X x-1 (c) 1 x-1 (f) 1 3x X X- 1 1 + i X 1 3(x + 2) = 3x + 6 A 2 -5 -1 x-(x-l) X X 12. (a) (fog)(x) = |g(x)| = ^. (b) (f°g)(x) = ^ = ^^1-^ = ^^!-^ = ^^ = ^sog(x) = x+ 1. (c) Since (fog)(x) = Vg(x) = |x|, g(x) = x 2 . (d) Since (fog)(x) = f(\/x) = |x |, f(x) = x 2 . (Note that the domain of the composite is [0, oo).) The completed table is shown. Note that the absolute value sign in part (d) is optional. g(x) f(x) (f°g)(x) 1 x-1 W 1 |x-l| X+l x-1 X X X+l X 2 ^ |x| V^ X 2 |x| 13. (a) f(g(x)) = ^[TT g(f(x)) l /x+l (b) Domain (fog): (0, oo), domain (gof): (— 1, oo) (c) Range (fog): (1, oo), range (gof): (0, oo) 14. (a) f(g(x)) = 1-2-y/x + x g(f(x)) = 1 - |x| (b) Domain (fog): (0, oo), domain (gof): (0, oo) (c) Range (fog): (0, oo), range (gof): (— oo, 1) 15. (a) y = -(x + 7) 2 16. (a) y = x 2 + 3 (b) y = -(x - 4) 2 (b) y = x 2 - 5 17. (a) Position 4 (b) Position 1 (c) Position 2 (d) Position 3 18. (a) y = -(x-l) 2 + 4 (b) y = -(x + 2) 2 + 3 (c) y=-(x + 4) 2 -l (d) y = -(x - 2) 2 Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Section 1.5 Combining Functions; Shifting and Scaling Graphs 29 19. 20. (* + 4)'t(y-3) , -25 y \ ^*- "^^ B I / \ 1 ii +)r -25 H / t \-6--4 -2/ 2 i ! x \ ; / / 21. 22. 23. 25. ? y = 2x 1 2 NX 1 2a ^^> -3 -2 -1 V i ,/'* ' -1 u,-i\ -2 \ 20 y + l-(x-l) 24. 26. Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 30 27. 29. Chapter 1 Preliminaries 28. j :4 :3 I / l/l IT ; 1 V. -6 -y - 2 V 2 4 6 .. . M -2 y + 1 = (x + 2) 30. 31. 32. 2 4 6 33. 34. 35. 3 2 1 - *(i. i) 1+1 /jc-1 1 2 5 36. Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 37. Section 1.5 Combining Functions; Shifting and Scaling Graphs 31 38. 39. 40. 41. 42. 1 2 " ; v = 1 x-1 1 j __0 -1 2 3 4 -2 - \ 7.5 5 2.5 l-H -4 -2 -2 \^2 4 ^2.5 43. y = - x +l 12 3 44. -7 y = X + 2 2 4 Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 32 45. Chapter 1 Preliminaries 1 . 4 -| ; l i 3 -/ ; 1 c»-i) 2 U -1 1 2 3 46. 1 ) 5 / 1 '-7-' _/j p — • ^-~? & -1 -5 -10 47. j « p / _ 1 _-' A - -2 - 1 1 2 48. • -5 y = (« + D 8 2 4 49. (a) domain: [0,2]; range: [2,3] > 3 \ y=/W + 2 2 1 2 3 4 (c) domain: [0,2]; range: [0,2] (b) domain: [0,2]; range: [-1,0] y = /W-l (d) domain: [0,2]; range: [—1,0] y = 2f(x) (e) domain: [—2,0]; range: [0,1] » t < 2 y=f(pc + 2) .f: 1 -2 -1 y=-m (f) domain: [1,3]; range: [0,1] > 2 1 l/ -1) 1 2 3 Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Section 1.5 Combining Functions; Shifting and Scaling Graphs (g) domain: [—2,0]; range: [0,1] (h) domain: [— 1, 1]; range: [0,1] 33 y=f(-x) y=-f(x + l) + l 50. (a) domain: [0,4]; range: [-3,0] (b) domain: [-4,0]; range: [0,3] «-t (c) domain: [—4,0]; range: [0,3] (d) domain: [-4,0]; range: [1,4] y = l -g(t) -2 (e) domain: [2,4]; range: [—3,0] y g(-t + 2) (f) domain: [-2,2]; range: [-3,0] Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 34 Chapter 1 Preliminaries (g) domain: [1,5]; range: [—3,0] 51. y = 3x 2 -3 52. y = (2x) 2 - 1 = 4x 2 - 1 53- y=K 1 +i)=l+2? 54 -y = 1 + ^? = 1 + p 55. y = V4X+1 56. y = 3v/x+ 1 57- y = \/4 -(f) 2 = |v / T6^ 58. y= i\/4~ 59. y = 1 - (3x) 3 = 1 - 27x 3 60. y = i- (|) 3 = i- (h) domain: [0,4]; range: [0,3] y i 3 \ y--g(t-4) 2 4 61. Lety = - V / 2x+l = f(x) and let g(x) = x 1 / 2 , h(x) = (x + 5) 1/2 ,i(x) = a/2(x + i) 1/2 ,and j(x) = — v 2(x +5) = i(x). The graph of h(x) is the graph of g(x) shifted left | unit; the graph of i(x) is the graph of h(x) stretched vertically by a factor of y 2; and the graph of j(x) = f(x) is the graph of i(x) reflected across the x-axis. Copffigl (c) 1 Pearson M% } k, publishing as Pearson Addison-Wesle Section 1.5 Combining Functions; Shifting and Scaling Graphs 35 62. Let y = ^/T^\ = f(x). Let g(x) = (~x) 1/2 , h(x) = (-x + 2) 1/2 , andi(x) = 4- (-x + 2) 1/2 = ^l^~\ = f(x). The graph of g(x) is the graph of y = ^fx reflected across the x-axis. The graph of h(x) is the graph of g(x) shifted right two units. And the graph of i(x) is the graph of h(x) compressed vertically by a factor of y 2. -3 -2-1 12 63. y = f(x) = x 3 . Shift f(x) one unit right followed by a shift two units up to get g(x) = (x — 1) +2. 64. y = (1 - xf + 2 = -[(x - l) 3 + (-2)] = f(x). Let g(x) = x 3 , h(x) = (x - l) 3 , i(x) = (x - l) 3 + (-2), and j(x) = — [(x — 1) + (—2)]. The graph of h(x) is the graph of g(x) shifted right one unit; the graph of i(x) is the graph of h(x) shifted down two units; and the graph of f(x) is the graph of i(x) reflected across the x-axis. g(x) h(x) 10 /y = x 5 -3 -2 -5 -10 1 2 3 10 y = (x-l) 3 / 5 J -3 -2 -1 / / 5 12 3 '-10 Copffigl (c) 1 Pearson %$A\ k, publishing as Pearson Addison-Wesle 36 Chapter 1 Preliminaries 65. Compress the graph of f(x) = j horizontally by a factor of 2 to get g(x) = ^- . Then shift g(x) vertically down 1 unit to geth(x) = i.-l. 66. Letf(x) = i and g(x) = £ + 1 = * + 1 — ^—2 + 1 = j-. — 1 2 + 1- Since v/2 « 1.4, we see that the graph of f(x) stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of g(x). 67. Reflect the graph of y = f(x) = ^fk across the x-axis to get g(x) = — -y/x. > 4 - 3 - 2 - -4 -3 -2 -1 -1 1 2 3 4 -2 -3 y = -fc -4 68. y = f(x) = (-2x) 2/3 = [(-l)(2)x] 2 / 3 = (-l) 2/3 (2x) 2/3 = (2x) 2/3 . So the graph of f(x) is the graph of g(x) = x 2 / 3 compressed horizontally by a factor of 2. f(x) Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle 69. Section 1.5 Combining Functions; Shifting and Scaling Graphs 70. 37 -3-2-1 12 3 71. 9x 2 + 25y 2 = 225 => h + h = 1 72. 16x 2 + 7y 2 = 112 =>• (Vr) 2 I 42 > v 6 - 4 9x 2 i 25/ = 225 2 -6 U4 -2 -2 4 -6 2 4 y 6 73. 3x 2 + (y - 2) = 3 => (y-2) 2 74. (x+l) 2 + 2y 2 = 4^ *-C-i). + W 4 " 3x 2 + (y - 2) 2 = 3 / 3 1 2 \ 1 -2 - 1 1 2 (x + l) 2 +2y 3 = 4 75. 3(x- l) 2 + 2(y + 2) 2 = 6 (x-i) 2 [y-(-2)] 2 _-, 76. 6(x+f) 2 + 9(y-i) 2 = 54 [-(-§)] , (y-ff ! 3 2 W 6(x + f) +9(y-^)"=54 3(j:-ir + 2(y + 2) z = 6 Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 38 Chapter 1 Preliminaries 2 2 77. fg + \ = 1 has its center at (0, 0). Shiftinig 4 units left and 3 units up gives the center at (h, k) = (—4, 3). So the equation is — — ^-^ — h " 32 ' from (-8, 3) to (0,3). (*- + ±Y , (y - 3)^ 42 -1- 32 1. Center, C, is (—4, 3), and major axis, AB, is the segment > . 10 (x + 4) 2 (y-3) 2 g 16 9 . L.J -10 -8 -6 -4 -2 2 -2 78. The ellipse \ + £ = 1 has center (h, k) = (0, 0). Shifting the ellipse 3 units right and 2 units down produces an ellipse with center at (h, k) = (3, —2) and an equation tx 4 ; + ^ — ^ r ^ 1. Center, C, is (3, —2), and AB, the segment from (3, 3) to (3, —7) is the major axis. 3 ' ( x-3) : _(y + 2f 3 2 1 4^-^25 -1 -1 -2 -3 -4 -5 -6 -7 2 3 4 )p 79. (a) (fg)(-x) = f(-x)g(-x) = f(x)(-g(x)) = -(fg)(x), odd (b) ( f -) (-x) = ^ = -SL = _ ( i) (X ), dd (O (f)(^) = f^ = li } = -(f)W'°dd gx) -g(x) -g(x) _ f(x) " (d) f 2 (-x) = f(-x)f(-x) = f(x)f(x) = f 2 (x), even (e) g 2 (-x) = (g(-x)) 2 = (-g(x)) 2 = g 2 (x), even (f) (f o g)(-x) = f(g(-x)) = f(-g(x)) = f(g(x)) = (f o g)(x), even (g) (g o f)(-x) = g(f(-x)) = g(f(x)) = (g o f)(x), even (h) (f o f)(-x) = f(f(-x)) = f(f(x)) = (f o f)(x), even (i) (g o g)(-x) = g(g(-x)) = g(-g(x)) = -g(g(x)) = -(g o g)(x), odd 80. Yes, f(x) = is both even and odd since f(— x) = = f(x) and f(— x) = -f(x). Copyright (c) 2006 Pearson Education Section 1.6 Trigonometric Functions 39 81. (a) (b) (c) (d) 82. (fog)(x) f(x) = x- 7 1.6 TRIGONOMETRIC FUNCTIONS 1. (a) s = r<9 = (10)(f ) =87rm 2. 6 = s - = ±2s = fK radians and 5 -f f 1 ^) = 225° r 8 4 4 \ n / (b) S = r0 = (10)(110°) (^) HOtt _ 55ir 18 9 111 3. 6> = 80° => 6> = 80° (^p) = f s = (6) (t^) = 8.4 in. (since the diameter = 12 in. =>- radius = 6 in.) Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle 40 Chapter 1 Preliminaries 4. d = 1 meter => r = 50 cm => 6> = f = | = 0.6 rad or 0.6 (±f£) « 34° 6» — TT 2tt 3 7T 2 3tt 4 sin x/3 2 1 1 7^ cos 1 1 2 1 1 tan 8 V^ und. -l cot 8 und. 1 75 und. -l sec 6* 1 -2 1 und. -V~2 esc 8 und. 2 und. 1 \pi 6. 6» 3tt 2 3 7T 6 7T 4 5tt 6 sin 6 1 1 2 1 2 1 75 1 2 cos 8 1 2 75 2 1 V~2 75 2 tan# und. ->/3 1 73 1 1 73 cot6> i -V3 1 -V* sec 8 und. 2 2 y5 V^ 2 75 esc 8 1 2 75 -2 V^ 2 7. cos x = — j, tan x = — | sinx= 7i' cosx= 75 9. sin x = — 7- , tan x = — y 8 10. sin x = j| , tan x = — y 11. sin x 1 2 -T- , COS X = t- 75 75 12. cos x 75 , tan x 75 13. period = 7r 14. period = 4tt 15. V ' * 3' = cos nx i 1 ° \ ' / 2 -1 period = 2 16. 1 o -l - y ' cos y ' 3tt n T period = 4 ... -\ y ■ cos x 17. period = 6 18. 1 t i... Ot -■ 1/ I 1 period = 1 y » - cos 2i« /TA A, "r2, ; \3 / \4/7"\5 y • cos x Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Section 1.6 Trigonometric Functions 41 19. period = 2ir 20. period = 2ir y sin (jt+|) 21. period = 2ir 22. \ 111 \ hi \ ! ^ -i\ period = 27r 23. period = |, symmetric about the origin 24. period = 1 , symmetric about the origin w lilt.. 111 fll \ II \ \! \ ! v 25. period = 4, symmetric about the y-axis 26. period = 4tt, symmetric about the origin s j 2^ -2n ! 2* -6|1 S * CSC 1 27. (a) Cos x and sec x are positive in QI and QIV and negative in QII and QUI. Sec x is undefined when cos x is 0. The range of sec x is (— oo, — 1] U [1, oo); the range of cos x is [— 1, 1]. y = cos x Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 42 Chapter 1 Preliminaries (b) Sin x and esc x are positive in QI and QII and negative in QUI and QIV. Csc x is undefined when sinx is 0. The range of csc x is (— oo, — 1] U [1, oo); the range of sin x is [— 1, 1]. y = csc x y y = sin x 28. Since cot x = — — , cot x is undefined when tan x = tan x and is zero when tan x is undefined. As tan x approaches zero through positive values, cot x approaches infinity. Also, cot x approaches negative infinity as tan x approaches zero through negative values. y » tan x 29. D: -oo < x < oo; R: y = -1, 0, 1 y = Lsin xj -2K ,* „-7T 1 L. k' h - h ' IH-^ > ' i >> y =f sin x| 31. cos (x — |) = cos x cos (— |) — sin x sin (— |) = (cos x)(0) — (sin x)(— 1) = sinx 32. cos (x + |) = cos x cos (|) — sin x sin (|) = (cos x)(0) — (sin x)(l) = —sin x 33. sin (x + |) = sin x cos (|) + cos x sin (|) = (sin x)(0) + (cos x)(l) = cos x 34. sin (x — | ) = sin x cos ( — f) + cos x sin ( — § ) = (sin x )(0) + (cos x)(— 1) = —cos x 35. cos (A - B) = cos (A + (-B)) = cos A cos (— B) - sin A sin (— B) = cos A cos B - sin A (-sin B) = cos A cos B + sin A sin B 36. sin (A - B) = sin(A + (— B)) = sin A cos (-B) + cos A sin(-B) = sin A cos B + cos A(-sinB) = sin A cos B — cos A sin B 37. If B = A, A - B = => cos (A - B) = cos 0=1. Also cos (A - B) = cos (A - A) = cos A cos A + sin A sin A = cos 2 A + sin 2 A. Therefore, cos 2 A + sin 2 A = 1 . 38. If B = 2tt, then cos (A + 27r) = cos A cos 2n — sin A sin 2n = (cos A)(l) — (sin A)(0) = cos A and sin (A + 2ir) = sin A cos 2ir + cos A sin 2ir = (sin A)(l) + (cos A)(0) = sin A. The result agrees with the fact that the cosine and sine functions have period 2ir. 39. cos (7T + x) = cos 7r cos x — sin 7r sin x = (— l)(cos x) — (0)(sin x) = —cos x Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Section 1.6 Trigonometric Functions 40. sin (2ir — x) = sin 2ir cos (— x) + cos (2tt) sin (— x) = (0)(cos (— x)) + (l)(sin (— x)) = —sin x 41. sin(^ -x) 42. cos(^+ X ; = sin (y) cos(— x) + cos (y) sin(— x) = (— l)(cos x) - = cos (y) cos x - sin (y ) sin x = (0)(cos x) — (— l)(sin x) (0)(sin = sin x 43. sin ^1 = sin (| 44. cos ^ =cos(| + f ) = sin | cos | + cos | sin y — sin | sin y (— x)) = —cos : 4 y") = cos f cos I ! = (f) (I) + (#)(#) = 45. cos fL=cos(f-f)=cosfcos(-|)-sinfsin(-|) = (i)(f)-(f)(-f) = i±f 46. sin 5| = sin (y' - |) = sin (^) cos (- |) + cos (^) sin I (*)(*) 43 47 Pnl; 2 tt _ l+cos(f) _ !+_# _ 2+y^ t/. cus g — 2 — 2 — 4 4 o rnQ 2 jr. _ 1+cos (!f ) _ 1 + 4 _ 2+^3 to. cus 12 — 2 — 2 — 4 4 n „ irl 2 jr. _ l-c°s(ff) _ \-& _ 2-y/3 *t7. sin 12 — 2 — 2 — 4 50. sin 2 f = i=^M = h± = ^f 51. tan(A + B) 52. tan (A - B) = sin( /^ = ^acosb-cos Acosg v y cos (A— B) cos A cos B+sin A so B tan A— tan B 1 +tan A tan B 53. According to the figure in the text, we have the following: By the law of cosines, c 2 = a 2 +-b 2 2ab cos 8 = 1 2 _|_ 1 2 _ 2 cos (A - B) = 2 - 2 cos (A - B). By distance formula, c 2 = (cos A - cos B) 2 + (sin A - sin B) 2 = cos 2 A — 2 cos A cos B + cos 2 B + sin 2 A — 2 sin A sin B + sin 2 B = 2 — 2(cos A cos B + sin A sin B). Thus c 2 = 2 - 2 cos (A - B) = 2 - 2(cos A cos B + sin A sin B) => cos (A - B) = cos A cos B + sin A sin B. 54. (a) cos(A — B) = cos A cos B + sin A sin B sin 9 = cos(f - 6>) and cos 9 = sin(f - 9) Let 9 = A + B sin(A + B) = cos If - (A + B)l = cos[(f - A) - b] = cos (§ - A) cos B + sin (§ - A) sin B = sin A cos B + cos A sin B (b) cos(A — B) = cos A cos B + sin A sin B cos(A - (— B)) = cos A cos (-B) + sin A sin (-B) => cos(A + B) = cos A cos (— B) + sin A sin (— B) = cos A cos B + sin A (—sin B) = cos A cos B — sin A sin B Because the cosine function is even and the sine functions is odd. 55. c 2 = a 2 + b 2 - 2ab cos C = 2 2 + 3 2 - 2(2)(3) cos (60°) = 4 + 9 - 12 cos (60°) =13-12 (|) = 7. Thus, c = s/l « 2.65. 56. c 2 = a 2 + b 2 -2abcosC = 2 2 + 3 2 - 2(2)(3)cos(40°) = 13- 12 cos (40°). Thus, c = sjll- 12 cos 40° « 1.951. Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 44 Chapter 1 Preliminaries 57. From the figures in the text, we see that sin B If C is an acute angle, then sin C g . On the other hand, if C is obtuse (as in the figure on the right), then sin C = sin (n — C) = \ . Thus, in either case, h = b sin C = c sin B => ah = ab sin C = ac sin B . By the law of cosines, cos C = u2 2 2 ° b ~ c and cos B 2_ b 2 2ac interior angles of a triangle is ir, we have sin A = sin (tt — (B (?) a 2 +b 2 -c 2 a 2 +c 2 -b 2 v 2abc ; (2a 2 + b 2 - c 2 Moreover, since the sum of the C)) = sin (B + C) = sin B cos C + cos B sin C 2 -b 2 ) = P => ah = be sin A. 2ab 2ac Combining our results we have ah = ab sin C, ah = ac sin B, and ah = be sin A. Dividing by abc gives _h be sin C sin B b law of sines 58. By the law of sines Thus sin B = ^ ~ 0.982. sin A sin B 2 — 3 '3/2 By Exercise 55 we know that c = yl. 59. From the figure at the right and the law of cosines, b 2 = a 2 +^ 2 2(2a) cos B 4-4a a 2 - 2a + 4. Applying the law of sines to the figure, sin B b ■/2Z2 y/3/2 b 2a + 4 = b 2 3 2 2 a | a. Thus, combining results, = 1 a 2 + 2a - 4 => = a 2 + 4a — 8. From the quadratic formula and the fact that a > 0, we have 60. (a) The graphs of y = sin x and y = x nearly coincide when x is near the origin (when the calculator is in radians mode), (b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves look like intersecting straight lines near the origin when the calculator is in degree mode. 61. A = 2,B = 2tt, C -7T,D 1 y = 2sin(x + n>- / n \ / 2 \ ' -1 % 2 / 3ff \ / 2 \ 5ir 2 -3 62. A= i,B = 2, C= 1,D -1 i y »-jSin (rtx - it) +i Copyright (c) 2006 Pearson Education Section 1.6 Trigonometric Functions 45 63. A 1,B = 4, C = 0,D y ,, y = 2 ■ ruts , l -5r sln lTj + 5? f\n 1 1' % -1 \ 1 / 3 \ 5 , X 64. A= ^,B = L, C = 0,D ^sm^.L>0 65. (a) amplitude = |A| = 37 (c) right horizontal shift = C = 101 (b) period = |B| = 365 (d) upward vertical shift = D = 25 66. (a) It is highest when the value of the sine is 1 at f(101) = 37 sin (0) + 25 = 62° F. The lowest mean daily temp is 37(— 1) + 25 = — 12° F. (b) The average of the highest and lowest mean daily temperatures = — ^- — - = 25° F. The average of the sine function is its horizontal axis, y = 25. 67-70. Example CAS commands: Maple f := x -> A*sin((2*Pi/B)*(x-C))+Dl; A:=3;C:=0;D1:=0; fjist := [seq( f(x), B=[l,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[ 1,3,4,7], legend=["B=l","B=3","B=2*Pi","B=3*Pi"], title="#67 (Section 1.6)"); Mathematica Clear[a, b, c, d, f, x] f[x_]:=a Sin[27r/b (x - c)] + d Plot[f[x]/.{a -»• 3, b ->■ 1, c -»■ 0, d 0}, {x, -47r,47r }] 67. (a) The graph stretches horizontally. B = 2 'Pi B = 1 B = 3 ; Pi Copyright (c) 2006 Pearson Education 46 Chapter 1 Preliminaries (b) The period remains the same: period | B |. The graph has a horizontal shift of \ period. 68. (a) The graph is shifted right C units. C = (b) The graph is shifted left C units. (c) A shift of ± one period will produce no apparent shift. | C | = 6 69. The graph shifts upwards | D | units for D > and down | D | units for D < 0. A TV > ' ,7V 7 V v / ,7 1 3 6- 4-_ \ 2-f > r\ • /~\ * '/ w 7 V / \* 1 \ ! \ ,D = 3 N / 7 \» x / 7 / - 10 \J- 5 V\ '/ -2 70. (a) The graph stretches | A | units. (b) For A < 0, the graph is inverted. 1.7 GRAPHING WITH CALCULATORS AND COMPUTERS 1-4. The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space. Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Section 1 .7 Graphing with Calculators and Computers 47 1. d. 2. c. y ffx) = x'-4x 2 -4x + 16 -5 -4 -3 -2 - 12 3 4 5 f(x) = x 4 -7x : +6x 3. d. 4. b. f(x) = 5 + 12x-x 3 5-30. For any display there are many appropriate display widows. The graphs given as answers in Exercises 5—30 are not unique in appearance. 5. [-2, 5] by [-15, 40] 6. [-4, 4] by [-4, 4] 7. [-2, 6] by [-250, 50] 8. [-1, 5] by [-5, 30] . ) /Ct)=x 5 -5x 4 +10 50 -2 / iV 3 4 f 6 / -50 / -100 / -150 / -200 J -250 30 . f(x) = 4x 3 -x 4 20 10 V 1 12 3' , 5 Coppgl (c) 1 Pen Etation, k, publishing as Pearson Addison-Wesle 48 Chapter 1 Preliminaries 9. [-4, 4] by [-5, 5] 11. [-2, 6] by [-5, 4] 13. [-1,6] by [-1,4] 15. [-3, 3] by [0,10] 10. [-2, 2] by [-2, e f(x) = x : (6-x 3 ) e /"\ 4 / \ \ 2 :/ \ -2 -1 i I2 -2 - L 1 12. [-4, 4] by [-8, 8 ' *\ 2 -4 y 3 -2 -i-2 1 2/3 4 -4 -6 -8 : 1/3 / 2 y = x (x - 14. [-1,6] by [-1,5] 16. [-1,2] by [0,1] -5-4-3-2-1 12 3 4 5 Coppgl (c) 1 Pen Etation, k, publishing as Pearson Addison-Wesle 17. [-5,1] by [-5, 5] Section 1 .7 Graphing with Calculators and Computers 18. [-5,1] by [-2, 4] 49 2 4 6 8 10 x + 3 19. [-4, 4] by [0, 3] 20. [-5, 5] by [-2, 2] 21. [-10, 10] by [-6, 6] 22. [-5, 5] by [-2, 2] -10-8 -6 -4~\ /(*) = -fr 23. [-6, 10] by [-6, 6] 24. [-3, 5] by [-2, 10] 1 1 a ^ 6. 2 - 15.+ 6 \ m= 4.2-10. "5 -2 -4 -6 5 10 >X Coppgl (c| 1 Pen Etation, k, publishing as Pearson Addison-Wesle 50 Chapter 1 Preliminaries 25. [-0.03, 0.03] by [-1.25, 1.25] v 1.0 / \ y = sin 250.* 0.5 1 \ -0.02 1 \ 0.02/ 27. [-300, 300] by [-1.25, 1.25] 1.0 ' v = cos (5o) -300 \ \ /300 >X 1-0.5 J -1.0 29. [-0.25, 0.25] by [-0.3, 0.3] 26. [-0.1, 0.1] by [-3, 3] y = 3cos60x 28. [-50, 50] by [-0.1, 0.1] 1 . f X y = — sin — 10 ho 30. [-0.15, 0.15] by [-0.02, 0.05] y = x" +;jcos lOOx 31. x 2 + 2x = 4 + 4y - y 2 =>• y = 2 ± a/-x 2 - 2x - The lower half is produced by graphing y = 2 - \/-x 2 - 2x + 8. (x+lf + (y-2) z = 9 32. y 2 — 16x 2 = 1 => y = ± \/l + 16x 2 . The upper branch is produced by graphing y = \J 1 + 16x 2 . Coppgl (c) 1 Pen %$A\ k, publishing as Pearson Addison-Wesle Section 1.7 Graphing with Calculators and Computers 51 33. ' /(x)=-tan2x 34. 10 f \ f(x) = 3cot|~ |+1 6 4 2 35. f f(x) = sin 2x + cos 3x 36. -6 -4 \-2 f ( x ) = sin 3 x 2 \4 /6 37. 4 3 • 2 1 »•*••••„ ~~~. 2 3 4 5 6 -1 •. -2 • -3 • y = lT3 -4 38. * •! -0.4. -0.2 !•-■ y = sm ■ f 1 sin — l x 0.2 • 0.4 39. 40. 1 8 . \ 1 y=x [x\ 6 t 5 .• \ 4 •■ \ 3 - * 2 / -5 -4 -3 -2 -1 -1 12 3 4 5 -2 x'-l 41. (a) y = 1059. 14x - 2074972 (b) m = 1059.14 dollars/year, which is the yearly increase in compensation. Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 52 Chapter 1 Preliminaries (c) 1985 1990 1995 2000 2005 (d) Answers may vary slightly, y = (1059. 14) (2010) - 2074972 = $53, 899 42. (a) Let C = cost and x = year. C = (7960. 71)x- 1.6 x 10 7 (b) Slope represents increase in cost per year (c) C = (2637.14)x - 5.2 x 10 6 (d) The median price is rising faster in the northeast (the slope is larger). 43. (a) Let x represent the speed in miles per hour and d the stopping distance in feet. The quadratic regression function is d = 0.0866x 2 - 1.97x + 50.1. (b) . . 500 400 300 200 100 20 40 60 80 100 (c) From the graph in part (b), the stopping distance is about 370 feet when the vehicle is 72 mph and it is about 525 feet when the speed is 85 mph. Algebraically: d quadratic (72) = 0.0866(72) 2 - 1.97(72) + 50.1 = 367.6 ft. d qU adratic(85) = 0.0866(85) 2 - 1.97(85) + 50.1 = 522.8 ft. (d) The linear regression function is d = 6.89x - 140.4 => d linear (72) = 6.89(72) - 140.4 = 355.7 ft and diinear(85) = 6.89(85) — 140.4 = 445.2 ft. The linear regression line is shown on the graph in part (b). The quadratic regression curve clearly gives the better fit. 44. (a) The power regression function is y = 4.44647x 0511414 . Copyright (c) 2006 Pearson Education Chapter 1 Practice Exercises 53 (b) 2 4 6 8 10 12 14 16 18 20 (c) 15.2 km/h (d) The linear regression function is y = 0.913675x + 4.189976 and it is shown on the graph in part (b). The linear regession function gives a speed of 14.2 km/h when y = 11 m. The power regression curve in part (a) better fits the data. CHAPTER 1 PRACTICE EXERCISES 1. 7 + 2x > 3 => 2x > -4 => x > -2 _i i i i j i -6-4-2 2 4 2. -3x< 10 => x > -f 10 3 ■4X 3. i(x-l) < ±(x-2) => 4(x- 1) < 5(x-2) =>4x-4<5x-10=^6<x x-3 > 4+x 3(x-3) > -2(4 + x) =4> 3x - 9 > -8 - 2x => 5x > 1 => x > | 5. |x+l| = 7=> x+ 1 = 7 or -(x 4-1) = 7 => x = 6orx 6. |y-3|<4^-4<y-3<4=^-l<y<7 -i—i — I I I I A I I I — +x 0123456789 7. |l-f|>§=> l-l<-|orl =>- x > 5 or x < — 1 x ^ 3 2^2 | < -§ or -| > i => -x < -5 or -x > 1 8. ^±? <5=> -5 < ^±Z < 5=^ _i5 < 2x4-7< 15=^ -22 < 2x < 8=^ -11 < x< 4 9. Since the particle moved to the y-axis, —2 4- Ax = => Ax = 2. Since Ay = 3Ax = 6, the new coordinates are (x 4- Ax, y 4- Ay) = (-2 4-2,5 4-6) = (0, 1 1). 10. (a) 10 - ■ • • 5 • • -10 -5 -5 -10 - • 5 10 Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle 54 Chapter 1 Preliminaries (b) line slope AB BC CD DA CE BD 2-8 -6 10-6 _ 4 _ 2 2 -(-4) — 6 — 3 6-(-3) _ _9_ _ 3 -4-2 -6 2 l-(-3) _ 4 _ 2 8-2 6 3 -4-f is vertical and has no slope (c) Yes; A, B, C and D form a parallelogram. (d) Yes. The line AB has equation y — 1 = — | (x — 8). Replacing x by y gives y = — 5 (t — 8) + 1 = — | (— y) + l=5+l=6. Thus, E (y .6) lies on the line AB and the points A, B and E are collinear. (e) The line CD has equation y + 3 = — § (x — 2) or y = — |x. Thus the line passes through the origin. 11. The triangle ABC is neither an isosceles triangle nor is it a right triangle. The lengths of AB, BC and AC are I / I q i V 53, v 72 and v 65, respectively. The slopes of AB, BC and AC are 5, — 1 and g, respectively. 12. P(x, 3x + 1) is a point on the line y = 3x + 1. If the distance from P to (0, 0) equals the distance from P to (-3, 4), then x 2 + (3x + l) 2 = (x + 3) 2 + (3 - 3x) 2 => x 2 + 9x 2 + 6x + 1 = x 2 + 6x + 9 + 9 - 18x + 9x 2 18x = 17 or x y = 3x + 1 = 3 (||) + 1 = ^. Thus the point is P '17 23 N . 18 ' 6 ) 13. y = 3(x - 1) + (-6) => y = 3x - 9 14. y = -i(x+l) + 2^y = -|x+ f 15. x = 16. m = -^_ 6 3] = =£■ = -2 => y = -2(x + 3) + 6 =» y = -2x 17. y = 2 18. m 5 - 3 -2 - 3 J2_ -5 -§^y=-f(x-3) + 3^y = -§x+f 19. y = -3x + 3 20. Since 2x — y = —2 is equivalent to y = 2x + 2, the slope of the given line (and hence the slope of the desired line) is 2. y = 2(x - 1) + 1 => y = 2x - 5 2 1 . Since 4x + 3y = 1 2 is equivalent toy = — fx + 4, the slope of the given line (and hence the slope of the desired line) is •f(x-4)-12^y iv _ 20 3 A 3 22. Since 3x — 5y = 1 is equivalent to y = |x — 4, the slope of the given line is | and the slope of the perpendicular line is -I- y=-!(* + 2)-3^y=-fx-f 23. Sinceix+ |y = 1 is equivalent to y = — |x + 3, the slope of the given line is — | and the slope of the perpendicular line is |. y = |(x + 1) + 2 => y = |x + | 24. The line passes through (0, —5) and (3, 0). m : 0-(-5) 5 3-0 3 => y = | X - 5 Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Chapter 1 Practice Exercises 55 25 . The area is A = 7r r and the circumference is C = 27r r. Thus, r=^=>A = 7r(;f- C\ 2 C 2 2- v 2?r / ' 4?r ' 26. The surface area is S = 4ttt 2 =>■ r = (jM . The volume is V = |tt r 3 =>■ r S \V2 Substitution into the formula for surface area gives S = 47rr = 4tt [^ 3V\ 2 / 3 27. The coordinates of a point on the parabola are (x, x 2 ). The angle of inclination 9 joining this point to the origin satisfies the equation tan 9 = — = x. Thus the point has coordinates (x, x 2 ) = (tan 9, tan 2 ^). 28. tan 9 = ^ = ^ => h = 500tan6> ft. run o u u 29. -3 -2 -1 y = x 12 3 30. Symmetric about the origin. Symmetric about the y-axis. 31. 32. y = x~ -2x-l Neither 33. y(-x 34. y(-x 35. y(-x 36. y(-x 37. y(-x 38. y(-x 39. y(-x Symmetric about the y-axis. (-x) 2 + 1 = x 2 + 1 = y(x). Even. (-x) 5 - (-x) 3 - (-x) = -x 5 + x 3 + x = -y(x). Odd. 1 — cos(— x) = 1 — cos x = y(x). Even. sec(— x) tan(— x) sin(-x) _ -sinx cos 2 (— x) cos 2 x -sec x tan x = — y(x). Odd. (-x) 4 +l _ x 4 +l x'+l (-x) 3 -2(-x) -x 3 +2x x 3 -2x -y(x). Odd. 1 — sin(— x) = 1 -I- sinx. Neither even nor odd. -x + cos(— x) = — x + cos x. Neither even nor odd. 40. y(-x) = V(- x ) 4 - ! = V 7 * 4 - 1 = y(x). Even. Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 56 Chapter 1 Preliminaries 41 42. 43. 44. 45. 46. 47. 50. 51. 52. The function is defined for all values of x, so the domain is (-co, oo). Since | x | attains all nonnegative values, the range is [—2, oo). Since the square root requires 1 — x > 0, the domain is [—oo, 1]. Since y 1 — x attains all nonnegative values, the range is [—2, oo). Since the square root requires 16 — x 2 > 0, the domain is [—4, 4]. For values of x in the domain, < 16 — x 2 < 16, so < y 16 — x 2 < 4. The range is [0, 4]. The function is defined for all values of x, so the domain is (— oo, oo). Since 3 2 ~ x attains all positive values, the range is (1, oo). The function is defined for all values of x, so the domain is (— oo, oo). Since 2e~ x attains all positive values, the range is (—3, oo). The function is equivalent to y = tan 2x, so we require 2x ^ y for odd integers k. The domain is given by x ^ ^f for odd integers k. Since the tangent function attains all values, the range is ( — oo, oo). The function is defined for all values of x, so the domain is (— oo, oo). The sine function attains values from —1 to 1, so —2 < 2sin(3x + 7r) < 2 and hence —3 < 2sin(3x + n) — 1 < 1. The range is [—3, 1]. The function is defined for all values of x, so the domain is (— oo, oo). The function is equivalent to y = y x 2 , which attains all nonnegative values. The range is [0, oo). The logarithm requires x — 3 > 0, so the domain is (3, oo). The logarithm attains all real values, so the range is (— oo, oo). The function is defined for all values of x, so the domain is (— oo, oo). The cube root attains all real values, so the range is (— oo, oo). The function is defined for —4 < x < 4, so the domain is [—4, 4]. The function is equivalent to y = y|x|, —4 < x < 4, which attains values from to 2 for x in the domain. The range is [0, 2]. The function is defined for — 2 < x < 2, so the domain is [—2, 2]. The range is [— 1, 1]. 53. First piece: Line through (0, 1) and (1, 0). m = Second piece: Line through (1, 1) and (2, 0). m 1 - x, < x < 1 2 - x, 1 < x < 2 o-i _ ^i _ 1-0 " 1 " 0-1 _ ^1 2-1 1 -l=^y=-x+l = l — x = -l=>y = -(x-l) + l -x+2 = 2 -x f(x) 54. First piece: Line through (0, 0) and (2, 5). m = Second piece: Line through (2, 5) and (4, 0). m 5-0 2-0 ~~ 2 - On! _ zj> ~~ 4-2 ~~ 2 ' ^ y = |x (x-2)+5 •|x+ 10= 10- f f(x) f x, < x < 2 (Note: x = 2 can be included on either piece.) 10-#, 2<x<4 Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle Chapter 1 Practice Exercises 57 55. (a) (b) (c) (d) 56. (a) (b) (c) (d) :og)(-l) = f(g(-l)) = f( 7 =TTl) = f (!) = T ,of)(2)=g(f(2))=g(i)= * 2.5 V 5 fof)(x)=f(f(x))=f(I) gog)(x) = g(g(x)) = g(;^=) i = x , x + l/x ' ' /x + 2 \/7^2 +2 \/l + 2V^+2 fog)(-l) = f(g(-l)) = f(-y=T+l) = f(0) =2-0 = 2 gof)(2) = f(g(2)) = g(2 - 2) = g(0) = ^OTT = 1 fof)(x) = f(f(x)) = f(2 - x) = 2 - (2 - x) = x °g)W = g(gW) = g(^Ti) = ^/^xTT + i 57. (a) (fog)(x) = f(g(x)) = f(Vx + 2) = 2 - (V^+2) = -x, x > -2. of)(x) = f(g(x)) = g(2 - x 2 ) = v / (2-x 2 ) + 2 = v/4-x 2 (b) Domain of fog: [—2, oo). Domain of gof: [—2, 2]. (c) Range of fog: (— oo, 2]. Range of gof: [0, 2]. 59. 58. (a) (fog)(x) = f(g(x)) = f(v / T^c) = TTr^t = </l (gof)(x)=f(g(x))=g(v^) = (b) Domain of fog: ( — oo, 1]. Domain of gof: [0, 1]. (c) Range of fog: [0, oo). Range of gof: [0, 1]. The graph of f2(x) = fi(|x|) is the same as the graph of fi(x) to the right of the y-axis. The graph of f2(x) to the left of the y-axis is the reflection of y = fi(x), x > across the y-axis. 60. The graph of f2(x) = fi (|x|) is the same as the graph of fi(x) to the right of the y-axis. The graph of f2(x) to the left of the y-axis is the reflection of y = fi(x), x > across the y-axis. Copyright (c) 2006 Pearson Education 58 61. Chapter 1 Preliminaries i It does not change the graph. 62. The graph of f2(x) = fi ( |x|) is the same as the graph of fi(x) to the right of the y-axis. The graph of f2(x) to the left of the y-axis is the reflection of y = fi(x), x > across the y-axis. 63. 64. The graph of f2(x) = fi(|x|) is the same as the graph of fi(x) to the right of the y-axis. The graph of f2(x) to the left of the y-axis is the reflection of y = fi(x), x > across the y-axis. The graph of f2(x) = fi (|x|) is the same as the graph of fi(x) to the right of the y-axis. The graph of f2(x) to the left of the y-axis is the reflection of y = fi(x), x > across the y-axis. 65. 66. \ y- 1* 3 | 1 I 7 y i .* i 1 .* '1 3 . y = x Whenever gi(x) is positive, the graph of y = g2(x) = |gi(x)| is the same as the graph of y = gi(x). When gi(x) is negative, the graph of y = g2(x) is the reflection of the graph of y = gi (x) across the x-axis. It does not change the graph. 67. Whenever gi(x) is positive, the graph of y = g2(x) = |gi(x)| is the same as the graph of y = gi(x). When gi(x) is negative, the graph of y = g2(x) is the reflection of the graph of y = gi(x) across the x-axis. Copyright (c) 2006 Pearson Education Chapter 1 Practice Exercises 59 68. Whenever gi(x) is positive, the graph of y = g2(x) = |gi(x)| is the same as the graph of y = gi(x). When gi(x) is negative, the graph of y = g2(x) is the reflection of the graph of y = gi(x) across the x-axis. 69. y = cos 2x / ° \ K I \ 2 / K \ 3ll I \ 2 / 2x ' 1 period = ir 70. -2u period = 4tt ysin-j 71. period = 2 72. period = 4 73. y i v 2 1, ;p = 2cos(x-|) Ik -1 3 5k\ An 6 \ 3 l\\K 1 6 -2 period = 2ir 74. period = 2-k y = 1 + sin tx+j) 75. (a) sin B = sin f = \ = \ => b = 2 sin f = 2 (^) = y^- By the theorem of Pythagoras, a 2 + b 2 = c 2 a= Vc 2 - b 2 = ^4-3 = 1. (b) sin B = sin ? _ b _ 2 3 c c (?y-vs ^ . Thus, a = Vc 2 - b 2 = ^(^) 2 -(2) 2 4 _ 2 5 " v^' 76. (a) sin A = - =4> a = c sin A 77. (a) tan B = ^ => a = -^ ^ ' a tan K (b) tan A (b) sin A a = b tan A Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 60 Chapter 1 Preliminaries 78. (a) sin A (c) sin A Vc2-b2 79. Let h = height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flatground, respectively. Then, tan 50° = ^, tan 35° = jj, and b - c = 10. C ' D ' Thus, h = c tan 50° and h = b tan 35° = => c tan 50° = (c + 10) tan 35° => c (tan 50° - tan 35°) = 10 tan 35° => c = , ' ' ari l 5 ° => h = c tan 50° (c+ 10) tan 35° tan 50°-tan 35° 10 tan 35° tan 50" ^ 1 1 no rn tan 50 —tan 35 fr — io -*r 80. Let h = height of balloon above ground. From the figure at the right, tan 40° = \, tan 70° = \, and a + b = 2. Thus, h = b tan 70° => h = (2 - a) tan 70° and h = a tan 40° => (2 - a) tan 70° = a tan 40° => a(tan 40° + tan 70°) = 2 tan 70° => a = , l!™ 70 ' => h = a tan 40° balloon tan 40°+tan 70° 2 tan 70° tan 40° ^ i -j i, *. Ana i .. ana '" S - J i-.J Jvlll. tan 40 + tan 70 81. (a) y 2t \y ,i y *sin x + cos: / (b) The period appears to be \-k. (c) f(x + 4tt) = sin (x + 4tt) + cos (^) = sin (x + 2tt) + cos (| + 2tt) since the period of sine and cosine is 27r. Thus, f(x) has period 47r. sin x cos 82. (a) y-s1n 7 (b) D = (-oo,0)U(0,oo);R= [-1,1] (c) f is not periodic. For suppose f has period p. Then f ( ^ + kp) =f(i) integers k. Choose k so large that ^- + kp > - f ( y + kp) = sin ( (1/2 x k J > which is a contradiction. Thus f has no period, as claimed. sin 2tt = for all < mk^v < n - Butthen Copyright (c) 2006 Pearson Education Chapter 1 Additional and Advanced Exercises CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 61 1. (a) The given graph is reflected about the y-axis. (c) The given graph is shifted left 1 unit, stretched vertically by a factor of 2, reflected about the x-axis, and then shifted upward 1 unit. y = -2f(x+l)+l (-4, 1) (b) The given graph is reflected about the x-axis. (-3, 0) (d) The given graph is shifted right 2 units, stretched vertically by a factor of 3, and then shifted downward 2 units. ? = 3/(x-2)-2 4L- »(2,4) (-1,-3) (-1,-2) (3,-2) 2. (a) (b) 3 « 2 » -3 7u/ \l/2 -3 -2 ■■ -3 3. There are (infinitely) many such function pairs. For example, f(x) = 3x and g(x) = 4x satisfy f(g(x)) = f(4x) = 3(4x) = 12x = 4(3x) = g(3x) = g(f(x)). 4. Yes, there are many such function pairs. For example, if g(x) = (2x + 3) 3 and f(x) = x 1 / 3 , then (f o g)(x) = f(g(x)) = f ((2x + 3) 3 ) = ((2x + 3) 3 ) 1/3 = 2x + 3. 5. If f is odd and defined at x, then f(— x) = — f(x). Thus g(— x) = f(— x) — 2 = — f(x) — 2 whereas - g (x) = -(f(x) - 2) = -f(x) + 2. Then g cannot be odd because g(-x) = -g(x) =>• -f(x) - 2 = -f(x) + 2 =S> 4 = 0, which is a contradiction. Also, g(x) is not even unless f(x) = for all x. On the other hand, if f is even, then g(x) = f(x) — 2 is also even: g(— x) = f(— x) — 2 = f(x) — 2 = g(x). 6. If g is odd and g(0) is defined, then g(0) = g(-0) = -g(0). Therefore, 2g(0) = =>• g(0) = 0. Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle 62 Chapter 1 Preliminaries For (x, y) in the 1st quadrant, |x| + |y| = 1 + x <=> x + y=l+x O y=l. For (x, y) in the 2nd quadrant, |x| + |y| = x + 1 <^> — x + y = x + 1 <=> y = 2x + 1. In the 3rd quadrant, |x| + |y| = x + 1 <& -x — y = x 4- 1 <^> y = -2x - 1. In the 4th quadrant, [x| + |y| = x + 1 -o- x + (— y) = x + 1 <=> y = — 1 . The graph is given at the right. |x| + \y\ =l+x 8. We use reasoning similar to Exercise 7. (1) 1st quadrant: y+jy|=x+|x| <^> 2y = 2x <^> y = x. (2) 2nd quadrant: y + |y| = x + x <^> 2y = x + (-x) = O y = 0. (3) 3rd quadrant: y + |y| = x + |x| & y + (-y) = x + (-x) <^> o = o =>• all points in the 3rd quadrant satisfy the equation. (4) 4th quadrant: y + |y| = x + jx| <^> y + (— y) = 2x <^> = x. Combining these results we have the graph given at the right: 9. By the law of sines, 10. By the law of sines, >/3 sin | sin A sin B a — b sin - ~ b~ sin A sin B sin B a — b — 3 y/3 sin (tt/4) _ v^ {^J sin (tt/3) VI lB = |sinf = |(f) 3\/2 8 11. By the law of cosines, a 2 = b 2 + c 2 — 2bc cos A =>• cos A b 2 + c 2 -a 2 _ 2 2 + 3 2 - 2 2 _ 3 2bc 2(2)(3) 12. By the law of cosines, c 2 = a 2 + b 2 - 2ab cos C = 2 2 + 3 2 - (2)(2)(3) cos £ = 4 + 9 - 12 (&\ = 13 - 6\fl =>■ c = v/l3 - 6-Jl, since c> 0. 13. By the law of cosines, b 2 = a 2 + c 2 — 2ac cos B => cos B a 2 +c 2 -b 2 _ 2 2 +4 2 -3 2 2ac (2)(2)(4) 4+16-9 16 U. Since < B < tt, sin B = \/ \ - cos 2 B = J\- i§ = v^p = ^ 16 16 14. By the law of cosines, c = a +b — 2ab cos C =4> cos C 2 +b 2 -c 2 _ 2 2 +4 2 -5 2 _ 4+16-25 2ab " (2)(2)(4) 16 - £. Since < C < tt, sin C = V T - cos 2 C = Jl - ^ = 4p. 256 ~~ 16 15. (a) sin 2 x + cos 2 x = 1 => sin 2 x = 1 — cos 2 x = (1 — cos x)(l + cos x) =4> (1 — cos x) l+COS X 1— cos x sin x sin x 1+cosx (b) Using the definition of the tangent function and the double angle formulas, we have tan 2 (f) s,„ 2 a) 2 cos 2 (J) " 1+C „ S ( 2 (|)) 1 —cos X 1 +COS X Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle Chapter 1 Additional and Advanced Exercises 63 16. The angles labeled 7 in the accompanying figure are equal since both angles subtend arc CD. Similarly, the two angles labeled a are equal since they both subtend arc AB. Thus, triangles AED and BEC are similar which implies ^ = 2ac a °^ b =>■ (a - c)(a + c) = b(2a cos 6 — b) ds e - b 2 2ab cos 9. 17. As in the proof of the law of sines of Section P. 5, Exercise 57, ah = be sin A = ab sin C = ac sin B =>■ the area of ABC = \ (base)(height) = i ah = i be sin A = i ab sin C = \ ac sin B. 18. As in Section P.5, Exercise 57, (Area of ABC) 2 \ (base) 2 (height) 2 - 4 1 Q 2u2 _ 1 „2u2 a^h a 2 b 2 sin J C a 2 + b 2 -c 2 I a 2 b 2 (1 — cos 2 C) . By the law of cosines, c 2 = a 2 + b 2 — 2ab cos C => cos C — — — Thus, (area of ABC) 2 = \ a 2 b 2 (1 - cos 2 C) = \ a 2 b 2 I 1 .I 2 + b 2 - c 2 2ab )*)-*{ 1 (a 2 + b 2 -c 2 ) 2N 4a 2 b 2 i (4a 2 b 2 - (a 2 + b 2 - c 2 ) 2 ) = i [(2ab + (a 2 + b 2 - c 2 )) (2ab - (a 2 + b 2 - c 2 ))] ^ [((a + b) 2 - c 2 ) (c 2 - (a - b) 2 )] = i [((a + b) + c)((a + b) - c)(c + (a - b))(c - (a - b))] [( a + b + c'l -a + b + c\ I a-b + c^ I a + b - c 2 2 > )] = s(s — a)(s — b)(s — c), where s a+b+c 2 Therefore, the area of ABC equals ^/s(s — a)(s — b)(s — c) . 19. 1. b + c — (a + c) = b — a, which is positive since a < b. Thus, a + c < b + c. 2. b — c — (a — c) = b — a, which is positive since a < b. Thus, a — c < b — c. 3. c > and a<b => c — = c and b — a are positive =>• (b — a)c = be — ac is positive =>• ac < be. 4. a < b and c < =>• b — a and — c are positive =^ (b — a)(— c) = ac — be is positive => be < ac. 5. Since a > 0, a and - are positive =4> - > 0. 6. Since < a < b, both i and 5 are positive. By (3), a < b and \ > => a (i) < b (j) or 1 < ^ ^ 1 (s)<a(5) b y(3)sincei>0 =* i<i. 7. a < b < =>■ i and 5 are both negative, i.e., \ < and 1 < 0. By (4), a < b and \ < => b Q) < a (^) ^ ;<l =► i(5)<Ms) b y(4)^ei<0 1 < i b V a 20. (a) If a = 0, then = |a| < |b| <^ b ^ o |a| 2 < Ibl 2 . Since lal 2 a 2 and b 2 we obtain a 2 < b 2 . If a ^ then |a| > and |a| < |b| ^> a 2 < b 2 . On the other hand, if a 2 < b 2 then a 2 |a| z < Ibl" 0<|b| (|b|-|a|)(|b| + |a|). Since (|b| + |a|) > and the product (|b| — |a|) (jbj + |a|) is positive, we must have (|b| — |a|) > =>• jbj > |a| . Thus |a| < |b| O a 2 < b 2 . (b) ab < |ab| => -ab > -2 |ab| by Exercise 19(4) above =^ a 2 - 2ab + b 2 > |a| 2 - 2 |a| |b| + |b| 2 , since |a| 2 = a 2 and |b| 2 = b 2 . Factoring both sides, (a - b) 2 > (|a| - |b|) 2 => |a-b| > ||a| - |b|| , by part (a). 21. The fact that |ai + a2 + ... + a n | < |ai| + |a.2 1 + ••• + |a„| holds for n = 1 is obvious. It also holds for n = 2 by the triangle inequality. We now show it holds for all positive integers n, by induction. Suppose it holds for n = k > 1: |ai + &2 + ■ ■ ■ + a k | < |ai| + |a2 1 + • ■ • + Kl (this is the induction hypothesis). Then |ai + a 2 + ... + a k + a t+1 [ = |(ai + a 2 + ... + a k ) + a k+1 j < jai + a 2 + ... + a k | + |a k+[ | (by the triangle inequality) < |ai | + |a 2 1 + . . . + |a k | + |a k+1 1 (by the induction hypothesis) and the inequality holds for n = k + 1 . Hence it holds for all n by induction. Cf igl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 64 Chapter 1 Preliminaries 22. The fact that |ai + a 2 + ... + a n | > |ai| — |&2 1 — ••• — |a n | holds for n = 1 is obvious. It holds for n = 2 by Exercise 21(b), since |ai + a 2 | = |ai - (-a 2 )| > ||ai| - |-a 2 || = ||ai| - |a 2 || > |ai| - |a 2 | . We now show it holds for all positive integers n by induction. Suppose the inequality holds for n = k > 1. Then |ai + a 2 + • • • + a k | > |ai| — |a 2 | — . . . — |a k | (this is the induction hypothesis). Thus |ai + ... + a k + a k+1 1 = |(ai + . . . + a k ) — (— a k+1 )| > ||(ai + ... +a k )|-|-a k+1 || (by Exercise 21(b)) = ||ai + ... +a k |-|a k+l || > |ai+... + a k |-|a k+l > |ai| — |a 2 | — ... — |a k | — |a k+1 | (by the induction hypothesis). Hence the inequality holds for all n by induction. 23. If f is even and odd, then f(— x) = — f(x) and f(— x) = f(x) =>• f(x) = — f(x) for all x in the domain of f. Thus 2f(x) = => f(x) = 0. 24. (a) As suggested, letE(x) = f(x) + f( ~ x) => E(-x) = «-») + 3-(-»» = f(x) + f( ~ x) = E(x) => E is an even function. Define O(x) = f(x) - E(x) = f(x) - f(x)+ 2 f( ~ x) = f(x) ~ f( ~ x) . Then O(-x) = f( - x) - 2 ( - ( - x)) = ^^ = - { M ^ 1 ) = -0(x) ^ O is an odd function => f(x) = E(x) + O(x) is the sum of an even and an odd function, (b) Part (a) shows that f(x) = E(x) + O(x) is the sum of an even and an odd function. If also f(x) = Ei(x) + Oi(x), where Ej is even and Oi is odd, then f(x) — f(x) = = (Ej(x) + Oi(x)) — (E(x) + O(x)). Thus, E(x) — E^x) = Oi(x) — O(x) for all x in the domain of f (which is the same as the domain of E — Ej and O — Oi). Now (E — Ei)(— x) = E(— x) — Ei(— x) = E(x) — Ei(x) (since E and Ej are even) = (E - Ej)(x) => E - Ei is even. Likewise, (Oi - 0)(-x) = Oi(-x) - O(-x) = -Oi(x) - (-O(x)) (since O and Oi are odd) = -(Oi(x) - 0(x)) = -(Oi - 0)(x) =4> Oi - O is odd. Therefore, E - Ej and Oi O are both even and odd so they must be zero at each x in the domain of f by Exercise 23. That is, Ei = E and Oi = O, so the decomposition of f found in part (a) is unique. 25. y = ax 2 +bx + c = a(x 2 + ^x+|j) - g + c = a (x + ^) 2 - f d + c (a) If a > the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward. If a < the graph is a parabola that opens downward. Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. (b) If a > the graph is a parabola that opens upward. If also b > 0, then increasing b causes a shift of the graph downward to the left; if b < 0, then decreasing b causes a shift of the graph downward and to the right. If a < the graph is a parabola that opens downward. If b > 0, increasing b shifts the graph upward to the right. If b < 0, decreasing b shifts the graph upward to the left. (c) Changing c (for fixed a and b) by Ac shifts the graph upward Ac units if Ac > 0, and downward —Ac units if Ac < 0. 26. (a) If a > 0, the graph rises to the right of the vertical line x = — b and falls to the left. If a < 0, the graph falls to the right of the line x = — b and rises to the left. If a = 0, the graph reduces to the horizontal line y = c. As |a| increases, the slope at any given point x = Xo increases in magnitude and the graph becomes steeper. As |a| decreases, the slope at Xo decreases in magnitude and the graph rises or falls more gradually. (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward. 27. If m > 0, the x-intercept of y = mx + 2 must be negative. If m < 0, then the x-intercept exceeds \ =>■ = mx + 2 and x > | =>■ x = - ^ > i =4> 0>m>-4. Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Chapter 1 Additional and Advanced Exercises 65 28. Each of the triangles pictured has the same base b = vAt = v(l sec). Moreover, the height of each triangle is the same value h. Thus h (base)(height) = \ bh = Ai = A2 = A3 = . . . . In conclusion, the object sweeps out equal areas in each one second interval. 'a+0 b+(p v 2 ' 2 I 'a M v2 ' 2) Thus the slope 29. (a) By Exercise #95 of Section 1.2, the coordinates of P are Ax a/2 a (b) The slope of AB = §5^ = — - . The line segments AB and OP are perpendicular when the product of their slopes is -1 = (2) (- \) = is perpendicular to OP when a = b. - K ■ Thus, b 2 a = b (since both are positive). Therefore, AB Copyright (c) 2006 Pearson Education 66 Chapter 1 Preliminaries NOTES: Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle CHAPTER 2 LIMITS AND CONTINUITY 2.1 RATES OF CHANGE AND LIMITS 1 . (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1 . There is no single number L that all the values g(x) get arbitrarily close to as x — > 1 . (b) 1 (c) 2. (a) (b) -1 (c) Does not exist. As t approaches from the left, f(t) approaches — 1. As t approaches from the right, f(t) approaches 1 . There is no single number L that f(t) gets arbitrarily close to as t — » 0. 3. (a) True (d) False (b) True (e) False (c) False (f) True 4. (a) False (d) True (b) False (e) True (c) True 5. lim A does not exist because A = - = lifx>0 and A = — = — 1 if x < 0. Asx approaches from the left, x > Q |x| |x| x |x| —x rr A approaches — 1 . Asx approaches from the right, A approaches 1 . There is no single number L that all the function values get arbitrarily close to as x — > 0. 6. As x approaches 1 from the left, the values of ^-r become increasingly large and negative. As x approaches 1 from the right, the values become increasingly large and positive. There is no one number L that all the function values get arbitrarily close to as x — > 1, so lim -^-r does not exist. X — » 1 x ' 7. Nothing can be said about f(x) because the existence of a limit as x — » xo does not depend on how the function is defined at Xo . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when x is close enough to Xo. That is, the existence of a limit depends on the values of f(x) for x near Xo, not on the definition of f(x) at x itself. 8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near regardless of x — > the value f(0) itself. 9. No, the definition does not require that f be defined at x = 1 in order for a limiting value to exist there. If f(l) is defined, it can be any real number, so we can conclude nothing about f(l) from lim f(x) = 5. X — > 1 10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(l) itself. If lim f(x) exists, its value may be some number other than f(l) X — > 1 We can conclude nothing about lim f(x), X — » 1 whether it exists or what its value is if it does exist, from knowing the value of f(l) alone. t(cl2 na. dud Bima as Dc 68 Chapter 2 Limits and Continuity 11. (a) f(x) = (x 2 -9)/(x + 3) X -3.1 -3.01 -3.001 -3.0001 -3.00001 -3.000001 f(x) -6.1 -6.01 -6.001 -6.0001 -6.00001 -6.000001 X -2.9 -2.99 -2.999 -2.9999 -2.99999 -2.999999 f(x) 5.9 The estimate is lim f(x) x — > -3 -5.99 -6. -5.999 -5.9999 -5.99999 -5.999999 (b) f(x) = (x 1 - 9)/(x + 3) (c) f(x) = £=% = (x+ x 3 f 3 " 3) = x - 3 if x / -3, and _ lim „ (x - 3) = -3 - 3 = -6. x^ -3 12. (a) g(x) = (x 2 - 2)1 (x - y/2) X 1.4 1.41 1.414 1.4142 1.41421 1.414213 g(x) 2.81421 2.82421 2.82821 2.828413 2.828423 2.828426 (b) (c) g(x) x^-2 gOc) = (x 1 - 2)/(x - V2) U-V2 a/2 if x ^ \/ 2 > and lim f x + \fl) = \Jl + \/2 = 2\/2. ti —* ,/9 V / ^ 13. (a) G(x) = (x + 6)/ (x 2 + 4x - 12) G(x) X -5.9 -5.99 -5.999 -5.9999 -5.99999 -5.999999 G(x) -.126582 -.1251564 -.1250156 -.1250015 -.1250001 -.1250000 X -6.1 -6.01 -6.001 -6.0001 -6.00001 -6.000001 -.123456 -.124843 -.124984 -.124998 -.124999 -.124999 Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesley Section 2.1 Rates of Change and Limits 69 (b) (c) G(x) 10 -10 -20 G(x) = (.x + 6)/(x 1 + ix-\2) x + 6 x + 6 (x 2 + 4x-12) (x + 6)(x-2) —^n if x ^ — 6, and lim — ^= x — 2 ' ' v , _c x — 2 x^ -6 -6-2 -0.125. 14. (a) h(x) = (x 2 -2x-3)/(x 2 -4x + 3) (b) h(x) X 2.9 2.99 2.999 2.9999 2.99999 2.999999 h(x) 2.052631 2.005025 2.000500 2.000050 2.000005 2.0000005 X 3.1 3.01 3.001 3.0001 3.00001 3.000001 1.952380 1.995024 1.999500 1.999950 1.999995 1.999999 y 10 i -l -» -10 -20 1 3 (c) h(x) = i-j x 7\ = ( rl\^ + \\ = ^y if x ^ 3, and lim *±i = Ui v/ v/ x 2 — 4x + 3 (x — 3)(x— 1) x— 1 ' ' X ^Q x— 1 3 — 1 15. (a) f(x) = (x 2 -l)/(|x|-l) X -1.1 -1.01 -1.001 -1.0001 -1.00001 -1.000001 f(x) 2.1 2.01 2.001 2.0001 2.00001 2.000001 X -.9 -.99 -.999 -.9999 -.99999 -.999999 f(x) 1.9 1.99 1.999 1.9999 1.99999 1.999999 (b) f(x) = (x 1 - l)/(|x| - 1) Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesley 70 Chapter 2 Limits and Continuity (c) f(x) (x+lXx-1) x-1 (x+lXx-1) -(x+1) x+1, x > Oandx + 1 , and lim (1 - x) = 1 - (-1) = 2. 1 — x, x < and x^-1 x — > - 1 16. (a) F(x) = (x 2 + 3x + 2)/(2- |x|) X -2.1 -2.01 -2.001 -2.0001 -2.00001 -2.000001 F(x) 1.1 -1.01 -1.001 -1.0001 -1.00001 -1.000001 X -1.9 -1.99 -1.999 -1.9999 -1.99999 -1.999999 F(x) -.99 -.999 -.9999 -.99999 -.999999 (b) FM = t>r 2 + 3j: + 2V(2-W) (C) F(X) x >0 rx + ^x + n > and lim (X+ 1) = -2+ 1 = -1. (x + 2)(x + i) _ x+1 x < Qandx ^ -2 * -» -2 i7. (a) g(0) = (sin eye e .i .01 .001 .0001 .00001 .000001 g(0) .998334 .999983 .999999 .999999 .999999 .999999 e -.1 -.01 -.001 -.0001 -.00001 -.000001 m .998334 .999983 .999999 .999999 .999999 .999999 lim g(0) = 1 (b) L — ,1^ 3 v- sin , ,. , \ y = — — (radians) \ \l U— \l L .1 -5tt -Ait =3tt -2)h Sir TT^-^27T 37T 4-7T 57T NOT TO SCALE 18. (a) G(t) = (l -cost)/t 2 t .1 .01 .001 .0001 .00001 .000001 G(t) .499583 .499995 .499999 .5 .5 .5 t -.1 -.01 -.001 -.0001 -.00001 -.000001 G(t) .499583 .499995 .499999 .5 .5 .5 lim G(t) = 0.5 t-t0 Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesley Section 2.1 Rates of Change and Limits 7 1 (b) H 1 H G(t) 1 - cos t 0.4 4-0.3 0.2 0.1 -H 1 1 -0.0OO3 -0.0001 0.0001 0.0003 Graph is NOT TO SCALE 19. (a) f(x) = x 1 /' 1 ^) X .9 .99 .999 .9999 .99999 .999999 f(x) .348678 .366032 .367695 .367861 .367877 .367879 X 1.1 1.01 1.001 1.0001 1.00001 1.000001 f(x) .385543 .369711 .368063 .367897 .367881 .367878 lim f(x) w 0.36788 X — > 1 (b) C.9999 0.99995 2.71815 Graph is NOT TO SCALE. Also the intersection of the axes is not the origin: the axes intersect at the point (1,2.71820). 20. (a) f(x) = (3 X - l)/x X .1 .01 .001 .0001 .00001 .000001 f(x) 1.161231 1.104669 1.099215 1.098672 1.098618 1.098612 X -.1 -.01 -.001 -.0001 -.00001 -.000001 f(x) 1.040415 1.092599 1.098009 1.098551 1.098606 1.098611 lim f(x) w 1.0986 x — > (b) 21. lim 2x = 2(2) = 4 x^2 22. lim 2x = 2(0) = x — » 23. lim i (3x-l) = 3(|) -1=0 24. lim ^-r = -,,,, 1 , x _> ] 3x-l 3(1)- 1 Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesley 72 Chapter 2 Limits and Continuity 25. lim 3x(2x - 1) = 3(-l)(2(-l) - 1) X — > — 1 26. lim &= ##, = -i x _» _J 2x— 1 2(— 1)— 1 -3 27. lim x sin x = I sin I = 5 Y , 7T Z Z 2 28. lim f^ = <pJi = _=L = i X — > 7T 1— 7T 1 — 7T l—TT 7T — 1 29. (a) M = f(^|2) = 28_9 = 19 CM Af _ f(l)-f(-l) _ 2^0 _ -, {"' Ax ~~ l-(-l) ~ 2 ~ l in |" a ) A S - gci)-g(-i) _ 1-1 _ n CM ^£ — g( Q )-g(- 2 > _ 0^4 _ _ 9 w Ax ~~ 0-(-2) ~ 2 ~ z - 31. (a) Ah _ h(f)-Mf) _ -1-1 At fU) Ah _ h(f)-h(f) _ Q-yf _ -3^3 1 ' At - § - f - f - ,r 32 (a) ^ = g (7r >~gW = (2-D-(2+D = _ 2 ' ^ ^ At 7T — 7T — 7T CM Ag _ g(7r)-g(-7r) _ (2-l)-(2-l) _ n W At ~~ TT-(-TT) — 271 _ U oo AR _ R(2) - R(Q) JJ - AS 2-0 8+1- \/T 3-1 2 o 4 AP _ P(2)-P(l) J ^' A0 2-1 (8 — 16 + 10)— (1 -4 + 5) 1 2-2 = 35. (a) Slope of PQ At Qi(10,225) Q 2 (14,375) Q 3 (16.5,475) Q 4 (18,550) 650 - 225 20- 650- 10 375 20-14 650-475 16.5 -550 20- 650 42.5 m/sec 45.83 m/sec 50.00 m/sec 50.00 m/sec (b) At t = 20, the Cobra was traveling approximately 50 m/sec or 180 km/h. 36. (a) Slope of PQ Ap At Qi(5,20) Q 2 (7,39) Q 3 (8.5,58) Q 4 (9.5,72) SO- -20 10 -5 so- -39 ld -7 80- -58 10- -S.5 80- -72 10-9.5 12 m/sec 13.7 m/sec 14.7 m/sec 16 m/sec (b) Approximately 1 6 m/sec 37. (a) 91 92 93 Year 174- -62 112 1994- 1992 2 56 thousand dollars per year (c) The average rate of change from 1991 to 1992 is Ap 62-27 At The average rate of change from 1992 to 1993 is— 1992-1991 1 1 1 - 62 35 thousand dollars per year. 'At | iw; _ |,„i — 49 thousand dollars per year. So, the rate at which profits were changing in 1992 is approximatley J (35 + 49) = 42 thousand dollars per year. Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesley Section 2.1 Rates of Change and Limits 73 38. (a) F(x) x (x + 2)/(x - 2) 1.2 1.1 1.01 1.001 1.0001 1 F(x) -4.0 -3.4 4 -°- ( 3 ' 5.0; -4.04; -4.0004 -3.04 AF Ax AF Ax Ax 1.2-1 — -3.04 -(-3) _ 1.01-1 -3.0004 -(-3) 1.0001- 1 (b) The rate of change of F(x) at x = 1 is —4. AF Ax AF Ax 3.004 -3.0004 -3 - - - 3 '^ ( - 3) - -4.4; 1.1 - 1 -3.004- (-3) 1.001- 1 -4.004; 39. (a) g = S^iffi 2-1 g(l+h)- gd) N/2-1 2- 1 v/l+h-1 0.414213 Ax (l+h)-l (b) g(x) = v /x Ag Ax (1-5) -g(l) 1 .5 - 1 1.5-1 0.5 0.449489 1+h 1.1 1.01 1.001 1.0001 1.00001 1.000001 \/l+h 1.04880 1.004987 1.0004998 1.0000499 1.000005 1.0000005 (Vl+h- - l)/h 0.4880 0.4987 0.4998 0.499 0.5 0.5 (c) The rate of change of g(x) at x = 1 is 0.5. 'l+h-1 1 (d) The calculator gives lim h^0 40. (a) i) f(3)-f(2) 3-2 ii) ^P I _ 1 T 2 T-2 21 2T __ ^— 1 T-2 2T(T - 2) 2-T -2T(2-T) _ J_ X ^ 2 2T' L r * (b) (c) The table indicates the rate of change is —0.25 at t = 2. T 2.1 2.01 2.001 2.0001 2.00001 2.000001 f(T) 0.476190 0.497512 0.499750 0.4999750 0.499997 0.499999 (f(T) - - f(2))/(T - -2) -0.2381 -0.2488 -0.2500 -0.2500 -0.2500 -0.2500 (d) lim T^ 2 J-) -2TV 41-46. Example CAS commands: Maple : f := x -> (x A 4 - 16)/(x - 2); xO := 2; plot( f(x), x = x0-l..x0+l, color = black, title = "Section 2.1, #4 1(a)" ); limit( f(x), x = xO ); In Exercise 43, note that the standard cube root, x A (l/3), is not defined for x<0 in many CASs. This can be overcome in Maple by entering the function as f := x -> (surd(x+l, 3) — l)/x. Mathematica : (assigned function and values for xO and h may vary) Clear[f, x] f[x_]:=(x 3 - x 2 - 5x - 3)/(x + l) 2 x0=-l;h= 0.1; Plot[f[x],{x,x0-h,x0 + h}] Limit[f[x],x — > xO] Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesley 74 Chapter 2 Limits and Continuity 2.2 CALCULATING LIMITS USING THE LIMIT LAWS 1. lim (2x + 5) = 2(-7) + 5 = -14 + 5 = -9 x — > — 7 2. lim (10 -3x)= 10-3(12)= 10-36= -26 3. lim (-x 2 + 5x - 2) = -(2) 2 + 5(2) -2= -4+ 10 -2 = 4 4. lim (x 3 - 2x 2 + 4x + 8) = (-2) 3 - 2(-2) 2 + 4(-2) 8 = -16 5. lim 8(t - 5)(t - 7) = 8(6 - 5)(6 - 7) = -8 t — ► 6 6. lim 3s(2s - 1) = 3 (|) [2 (f) - l] = 2 (f - f 7. lim x^2 x + 3 _ 2 + 3 _ 5 x + 6 ~~ 2 + 6 ~~ 8 8 - ^5 x^7 = 5^7 = -2 = "2 y^-5 ■ 2 (-5) 2 _ u 5 5-y 5 -(-5) ~ 10 _ 2 9 . urn X- - -<=£- -25-5 10 lim y + 2 - 2 + 2 _ 4 A - i y ^2 y 2 + 5y + 6 (2)2 + 5(2) + 6 4+10 + 6 20 5 11. lim 3(2x- l) 2 = 3(2(-l)- l) 2 = 3(-3) 2 = 27 X — > — 1 12. lim (x + 3) 1984 = (-4 + 3) 1984 = (-1) 1984 = 1 x — > — 4 13. lim (5-y) 4 / 3 = [5-(-3)] 4 / 3 = (8) 4 / 3 = ((8) 1 / 3 ) 4 = 2 4 = 16 y — > -3 v ' 14. lim (2z - 8) 1 / 3 = (2(0) - 8) 1 / 3 = (-8) 1 / 3 = -2 z — > 15. lim h^o \/3h+T+i V 3 <°) + 1 + 1 \A+i 3 _ 3 i(S lim 5 = 5 = 5 _ 5 h^O v / 5h + 4 + 2 v/5(0) + 4 + 2 ,/4 + 2 4 17. 1^ ^-i = lim h -► h h^O h y/jh + 1-1 y/3h+l + l V^+l = Um Ph+l^l = Hm 3h_ = Um V3h+1 + 1 h^0 h(y3h+l + lj h-»0 h^h+l + lj h -> /3h+l + l _J _ 3 /T + l 2 18. lim h->0 y/5h + 4-2 h lim h-»0 lim (5h + 4)-4 \/5h + 4-2 y/5h + 4 + 2 h V5h + 4 + 2 ~~ i/To h(v/5h + 4 + 2) lim 5h lim h( v /5h + 4 + 2) h^O \/5h + 4 + 2 /i + 2 4 19. lim x->5 x-5 x 2 -25 lim lim 5 (x + 5)(x-5) x *^ 5 x + 5 5+5 10 20. lim 2 *t 3 _^ x —> — 3 x +4x + 3 lim * + 3 _ lim -J— = I — - I x" -3 (x + 3)(x+l) x " -3 x + 1 -3 + 1 2 21. lim x^ -5 x + 5 lim x^ -5 (x + 5)(x-2) x + 5 lim (x - 2) = -5 - 2 = -7 -►-5 Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley Section 2.2 Calculating Limits Using the Limit Laws 75 22. lim x2 - 7x + 10 x^2 x - 2 lim. <»-/)(» -2) = lim (x - 5) = 2 - 5 = -3 x^2 x-2 x^ 2 23. lim t-> 1 tl+1^1 _ Hm (t + 2)(t-l) = lim t + 2 _ 1+2 _ 3 t 2 -l t_>l (t~D(t+l) t^i t+1 1 + 1 2 lim <™±1> lim t+2 _ -1+2 24 lim ' + 3t + 2 .,,,, ,,,,, t-T-l t'-t-2 ~~ t"-l (t-2)(t+l) ~~ t " "±i t-2 ~ -1-2 25. lim t^4 = lim =$£2 = lim ^ = ^ = -\ x _> _2 xJ + 2x 2 x^-2 x 2 (x + 2) x ^ -2 x 4 2 ?fi lim ^ ^^ .MM MM, Z °- y™0 3y'-16y2 ~ /To y 2 (3y 2 -16) ~ ^"\ 3y 2 - 16 " -16 lim lim 5y + 8 _ _8_ j im (u 2 + l)(u+l)(u-l) _ lim (u 2 + l)(u+l) _ (1 + 1X1 + 1) _ 4 27 Hm u l u"l u^-l-^-"! (u2 + u+l)(u-l) "u 1 ""! u2 + u + l lim (v - 2) (v 2 + 2v + 4) lim v 2 + 2v + 4 4 + 4 + 4 _ 12 _ 3 90 i; m v - 8 Z0 - v" l 2 v 4 - 16 _ y 1 :^ (v-2)(v + 2)(v2 + 4) ~ v ^" 2 ( v + 2)(v2 + 4) " (4)(8) "32 29. lim x^9 xA-3 lim xA-3 lim l x " 9 x^9 (v /x - 3 )(v^ + 3 ) x^9 v^ + 3 v^ + 3 6 x(2+y / x')(2-yx") 30. lim £=£ = lim ^x) = lim x ^ 4 2 - V x x^4 2 - x /x x ^ 4 2-, /x lim x (2 + a/x) = 4(2 + 2) = 16 31. lim x-l lim _^_^_^^ - Hm i^M±3 + 2) = Um (V^T3 + 2 ) X ->1 v^T+3-2 X _>1 (V^+3-2) (VxT3 + 2) x _> 1 (x + 3)-4 x _> j = x/4 + 2 = 4 32. lim V^T$-3 _ ,. (v / x^Ti-3)(7x2T8 + 3) x+1 lim lim (x 2 + 8)-9 1 (x + l)(Vx 2 + 8 + 3) x->-l (x+l)(V" 2 + 8 + 3) lim (x+l)(x-l) lim x-»-l (x+l)(Vx 2 + 8 + 3) x->-l \/x 2 + 8+3 3 + 3 33. lim ^- 4 = lim. (^-*)J_g^ + *) = lim. (x 2 +12)-16 x^2 lim x^2 x^2 (x-2)(x + 2) (x-2) (Vx2 + 12 + 4) x^2 (x-2)(Vx 2 + 12 + 4) lim x + 2 (x-2) (\7x 2 + 12 + 4) x^2 Vx 2 +12 + 4 \/l6 + 4 2 34. lim x + 2 2 V* 2 + 5 - 3 lim lim (x + 2)(v'x 2 + 5 + 3) (7x2 + 5-3) (Vx 2 + 5 + 3) (x + 2)(y^TI + 3) _ ^5 + 3 _ (x + 2)(x-2) lim x-2 = lim x^ -2 (x + 2)(Vx 2 + 5 + 3j (x 2 + 5) - 9 v/9 + 3 -4 3 2 35. lim 2 -vgp x^-3 x + 3 lim (2 - v/ X 2 - 5) (2 + \/x 2 - 5) 3 (x + 3)(2 + v/x 2 -5) x ^ -3 (x + 3)(2+ v / x 2 -5) lim 4-(x 2 -5) lim 9-x 2 3 (x + 3)(2 + Vx 2 -5) lim (3-x)(3 + x) lim 3 (x + 3)f2+^x 2 -5) x -> -3 2 + \/x 2 - 5 2 + v/4 Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley 76 Chapter 2 Limits and Continuity (4 - x) (5 + Vx2 + 9) . (4 - x) (5 + 7x2 + 9) 36. lim — 4 , * n = lim -, — . ' v N , "- . ' s = lim ^ — x -> 4 5 - Vx2 + 9 x -> 4 (5 - 7x2 + 9 ) (5 + \/x^+9 ) x -> 4 25 - (x 2 - lim x^4 (4 - x) (5 + Vx 2 + 9) (4 - x) (5 + \/x2 + 9) = lim ,. v ,,, . , — - = lim x -» 4 ( 4 ~ x )< 4 + x) x -> 4 16 -x 2 5 + ^x2 + 9 _ 5+y/25 _ 5 4 + x ~~ 8 ~ 4 37. (a (b (c 38. (a (b (c 39. (a (b (c (d 40. (a (b (c (d 41. (a (b (c (d 42. (a (b (c quotient rule difference and power rules sum and constant multiple rules quotient rule power and product rules difference and constant multiple rules x lim c f( X )g(x) = [ x lim c f(x)j [ x lim g(x)] = (5)(-2) = -10 x lim c 2f(x) g(x) = 2 [ x lim c f(x)] [ x lim c g(x)] = 2(5)(-2) = -20 lim [f(x) + 3g(x)] = lim f(x) + 3 lim g(x) = 5 + 3(-2) = -1 f(x) X — » c lim f(x) x — > c ' 5 x lml c f(x)-g(x) lim f(x) - lim g(x) 5-(-2) 7 lim [g(x) + 3] = lim g(x) + lim 3 = -3 + 3 = x ^ 4 x ^ 4 x ^ 4 lim xf(x) = lim x • lim f(x) = (4)(0) = x — > 4 x — > 4 x — > 4 lim [g(x)] 2 = f lim g(x)l 2 = [-3] 2 = 9 x — > 4 Lx — » 4 J lim -iW- = ^4 S(X) = ^1 = 3 v -I A f(x) - 1 lim f(x) - lim 1 0-1 J lim [f(x) + g(x)] = lim f(x) + lim g(x) = 7 + (-3) = 4 x — > b x^b x^b lim f(x) • g(x) = [ lim f(x)l [ lim g(x)l = (7)(-3) = -21 x^b Lx^b JLx^b J lim 4g(x) = f lim 4J [ lim g(x)l = (4)(-3) = -12 x — > b Lx^b J Lx^b J lim f(x)/g(x) = lim f(x)/ lim g(x) x — > b x^b x — » b J_ -3 lim [p(x) + r(x) + s(x)] = lim p(x) + lim r(x) + lim s(x) = 4 + + (-3) = 1 x — > — 2 x — > -2 x — > -2 x — » — 2 x lim 2 p(x) • r(x) • s(x) = [ lim ^ p(x)] [^ lim ^ r(x)] ^ lim ^ s(x)] = (4)(0)(-3) = lim [-4p(x) + 5r(x)]/s(x) = [-4 lim p(x) + 5 lim r(x)| / lim s(x) = [-4(4) + 5(0)]/- 3 = f 43. lim (l+h) 2 -l 2 h^O n h^0 (-2 + h)2-(-2)' lim 1+2h + h2 -' = lim ^±^ = lim (2 + h) = 2 h^0 h->0 44. lim h->0 lim 4 - 4h + h2 - 4 = lim ^^ = lim (h - 4) = -4 h^0 h h^O h h -f 45. lim [3(2 + h)- 4] -P(2) -4] = Hm 3h h^0 h->0 46. lim ( - 2 H ( - 2) h^O h ""' " Hm - 2 - ( - 2 + 1 ) = lim - h h^O 2h !>""') --'"-^ "' ~ h n h ' 4 -- h > Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley 47. lim /7+h- y/7 h lim h->0 Section 2.2 Calculating Limits Using the Limit Laws (v^Th-V^)(v^Th+V^) _ (7 + h)-7 77 h v / 7+h+V7 lim h^O hIJl + h- ■Vt) lim lim h->0 h(/7+h+/7) h->0 x/v+h+v/? 2s/l h->0 h h->0 h(v/3h+l+ 1J h-»0 h ^3h + 1 + 1 lim 3h lim 3 _ 3 h( v / 3h+l + l) h->0 \/3h +1 + 1 2 49. lim ^5 - 2x 2 = J 5 - 2(0) 2 = \/5 and lim v^ - x 2 = a/5 - (0) 2 = Jl; by the sandwich theorem, x ^ x^O lim f(x) = y^ x — ► 50. lim (2 — x 2 ) =2 — = 2 and lim 2 cos x = 2(1) = 2; by the sandwich theorem, lim g(x) = 2 x^O x^O x^O 51. (a) lim ( 1 — *H =1 — 2 = 1 and lim 1 = 1; by the sandwich theorem, lim = x - sin x = 1 x -> V 6 / 6 x -> J ' x _> 2-2 cos x (b) For x ^ 0, y = (x sin x)/(2 — 2 cos x) lies between the other two graphs in the figure, and the graphs converge as x — » 0. y = (x sin x)/(2 - 2 cos x) h(x) = 1 52. (a) lim I k ~ ta) = lim h — lim ^ = ^—0=4 and lim i = i ; by the sandwich theorem, x^0V 224 /x^0 2 x^0 242 2 x->0 2 2 •* lim i^f^ = 1 x -> x2 2 (b) For all x/0, the graph of f(x) = (1 — cos x)/x 2 lies between the line y = \ and the parabola 1 „2 x^/24, and the graphs converge as x — ► 0. 2 24 53. lim f(x) exists at those points c where lim x = lim x. Thus, c = c => c (1 — c) = X — » C X — > C ,2 c = 0, 1 , or — 1 . Moreover, lim f(x) = lim x = and lim f(x) = lim f(x) = 1 . x^0 x^O x ^ — 1 x^l 54. Nothing can be concluded about the values of f, g, and h at x = 2. Yes, f(2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f(x) =—5^0. x^2 ~ 55. 1 = lim x^4 ftv-i e lim f(x) — lim 5 lim f(x) — 5 ^= X Hmx-lmT2 = ^^T~ =* lim f(x) - 5 = 2(1) => lim f(x) = 2 + 5 = 7. X— *4 X )4 A ' '4: A f l ± Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley 78 Chapter 2 Limits and Continuity fix) 56. (a) 1 = lim -j x — > -2 x lim f(x) lim f(x) x^-2 _ x^-2 lim x 2 4 x^-2 lim f(x) = 4. x^ -2 (b) 1= lim x — > -2 x ffx) lim x^ -2 fix) lim 1 x^-2 x lim x^ -2 f(x) / l lim x^ -2 ffx) -2. 57. (a) = 3 - = [lim M=*\ [lim (x - 2)] = ^ [ f -f^- ) (x - 2) | = lim„ [f(x) - 5] = Jim„ f(x) - 5 x^2 x^2 lim ffx) = 5. x^ 2 (b) = 4 • = [ lim f -^5f} [ lim (x - 2)1 => lim f(x) = 5 as in part (a). La ' £ J La. ? Z. J X. f £* 58. (a) = 1 • = lim |£ lim x x -> x J Lx -> Inn ':)- ' lim x 2 1 = lim | Sg • x 2 lim f(x). That is, lim f(x) = 0. x — > x — > (b) = 1 • = [ lim ^1 [ lim xl = lim [^ - xl = lim ® . That is, lim ^ = 0. Lx -> x J Lx -> J x -> L x J x -> x x -> x 59. (a) lim x sin ! = v ' x -> x \ ly — x sin • 1 (b) -1 < sini < 1 forx^O: x > =>• —x < x sin - < x x < => —x > x sin ; > x lim x sin - = by the sandwich theorem; x^O x J lim x sin - = by the sandwich theorem. x^O x J 60. (a) lim x 2 cos (4) x — > vx (b) -1 < cos (^) < 1 forx ^ => -x 2 < x 2 cos (^) < x 2 theorem since lim x 2 = 0. x^O 2.3 PRECISE DEFINITION OF A LIMIT i. -i- ^-^ 0.4 /i(x) = x 2 cos(l/i 3 ) -0.4 lim x 2 cos (4) = by the sandwich x — » v x Step 1: \x-5\<6^>-6<x-5<6^>-6 + 5<\<5 + 5 Step 2: .5 + 5 = 7 =^> 6 = 2,or-6 + 5=l =5> 6 = 4. The value of S which assures x — 5| < <5 =$■ l<x<7is the smaller value, 6 = 2. Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesley Section 2.3 Precise Definition of a Limit 79 -+- -+X 1 2 7 Step 1: \x-2\<6=>-6<x-2<S=>-6 + 2<x<S + 2 Step 2: -6 + 2=1 => 6 = 1, or 6 + 2 = 1 => 6 = 5. The value of 6 which assures |x — 2| < 6 => l<x<7is the smaller value, 6=1. +- ->-^ -7/2 -3 -1/2 Step 1: |x - (-3)| <8 => -6 < x + 3 < 6 => -6-3<x<<5-3 Step 2: -6-3 = -|^>6=i,or6-3 = -i=^<5=|. The value of 6 which assures |x — (— 3)| < <5 => — |<x<— | is the smaller value, 6 = |. H ) »-x 2 2 Step 1: |x - (- |) | < 6 => -6 < x + | < 6 => -6 - § < x < <5 Step 2: -6 - | = - | => <5 = 2, or 6 - § = - \ => 6 = 1. The value of 6 which assures |x - 1)1 <« | < x < — | is the smaller value, 6 = 1. 5. -i, +^x 4/9 1/2 4/7 Step 1: |x-i|<(5=>-(5<x-5<6^>-i5+i<x<6+i Step 2: -6 + | = I => 6 = i, or 6 + ± = f => 6 = i The value of 6 which assures |x — || < 6 => | < x < | is the smaller value, 6 = X . -f +- -) •"« 2.7591 3 3.2391 Step 1: |x-3|<<5=^-6<x-3<<5=^-6 + 3<x<6 + 3 Step 2: -6 + 3 = 2.7591 => <5 = 0.2409, or 6 + 3 = 3.2391 => <5 = 0.2391. The value of 6 which assures |x — 3| < 6 =4> 2.7591 < x < 3.2391 is the smaller value, 6 = 0.2391. 7. Step 1: |x-5|<<5=^-6<x-5<<5^--6 + 5<x<<5 + 5 Step 2: From the graph, -6 + 5 = 4.9 =4> 6 = 0.1, or 8 + 5 = 5.1 => 6 = 0.1; thus 8 = 0.1 in either case. 8. Step 1: |x - (-3)| <8 => -8 < x + 3 < <5 => -6-3<x<<5-3 Step 2: From the graph, -6-3 = -3.1 => 6 = 0.1, or 6 - 3 = -2.9 => 8 = 0.1; thus 6 = 0.1. 9. Step 1: |x - 1| < 8 => -8 < x - 1 < 8 =>- -8+ 1 < x < 8 + 1 Step 2: From the graph, -8+1 = ^ => 8 = ^, or 6 + 1 = f| =>• 6 = ^; thus 6 = ^. 10. Step 1: \\~3\<8=>-8<\-3<8^-8 + 3<x<8 + 3 Step 2: From the graph, -6 + 3 = 2.61 => 6 = 0.39, or 8 + 3 = 3.41 => 8 = 0.41; thus 8 = 0.39. 11. Step 1: |x-2|<<5^-6<x-2<<5^-6 + 2<x<<5 + 2 Step 2: From the graph, -6 + 2=a/3 =4> 6 = 2 - a/3 « 0.2679, or 6 + 2 = a/5 => 6 = a/5 - 2 « 0.2361; thus 6 = a/5 - 2. Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley 80 Chapter 2 Limits and Continuity 12. Stepl: |x-(-l)|<<5 => -6<x+l<6 => -6-l<x<6-l Step 2: From the graph, —6—1 /5-2 0.1180, or 6- 1 = - A 2-V3 0.1340; thus (5 /5-2 13. Stepl: |x-(-l)|<<5 => -6<x+l<6 => -<5 - 1 < x < <S - 1 Step 2: From the graph, -(5 - 1 = - f => § = | « 0.77, or 6 - 1 - lh 25 "^ 25 ^ - 0.36; thus 5=4 = 0.36. 25 14. Step 1: |x- \\ < S =4> -6 < x - \ < 6 =^ Step 2: From the graph, — 6 + \ = jjjy =>- 6 thus 6 = 0.00248. -6+±<x<6+| 1 2.01 0.00248, or 6 + \ = ^ 6 = m~ I ~ 0.00251; 15. Step 1: |(x + 1) - 5| < 0.01 => |x - 4| < 0.01 => -0.01 < x - 4 < 0.01 => 3.99 < x < 4.01 Step 2: |x - 4| < 6 => -6 < x - 4 < 6 => -6+4<x<6+4 => 6 = 0.01. 16. Step 1: |(2x - 2) - (-6)| < 0.02 => |2x + 4| < 0.02 =^> -0.02 < 2x + 4 < 0.02 => -4.02 < 2x < -3. => -2.01 <x < -1.99 Step 2: |x - (-2)| < 6 => -6 < x + 2 < 6 => -6-2<x<6-2 => 6 = 0.01. 17. Step 1: k/x+ 1 - 1 < 0.1 =>• -0.1 < a/x + 1 - 1 < 0.1 => 0.9 < a/x + 1 < 1.1 => 0.81 <x+ 1 < 1.21 => -0.19 <x < 0.21 Step 2: |x - 0| < 6 => -6 < x < 6. Then, -5 = -0.19 ^ 6 = 0.19 or 6 = 0.21; thus, 6 = 0.19. 18. Step 1: | y/x - 1 1 < 0.1 => -0.1 < y/x - \ < 0.1 => 0.4 < ,/x < 0.6 => 0.16 < x < 0.36 Step 2: |x-±|<6=^-6<x-i<6=4--6+±<:r<6+i. Then, -6 + \ = 0.16 =^ 6 = 0.09 or 6 + \ = 0.36 ^ 6 = 0.11; thus 6 = 0.09. 19. Stepl: k/l9-x-3 < 1 =^ -1< a/19-x-3 < 1 => 2 < a/19-x <4 ^ -4 > x - 19 > -16 =^> 15>x>3or3<x<15 Step 2: |x - 10| < 6 => -6 < x - 10 < 6 => -6 + 10 < x < 6 + 10. Then -6+10 = 3 => 6 = 7, or 6 + 10 = 15 =4> 6 = 5; thus 6 = 5. => 4< 19-x< 16 20. Step 1: |a/x-7-4| < 1 =$► -1 < i/x - 7 - 4 < 1 => 3 < y/x - 7 < 5 => 9<x-7<25 => 16<x<32 Step 2: |x - 23 1 < 6 => -6 < x - 23 < 6 => -6 + 23 < x < 6 + 23. Then -6 + 23 = 16 =>■ 6 = 7, or 6 + 23 = 32 => 6 = 9; thus 6 = 7. 21. Step 1: |i - 1| < 0.05 => -0.05 < 1 - \ < 0.05 => 0.2 < ± < 0.3 = Step 2: |x - 4| < 6 =>• -6 < x - 4 < 6 => -6 + 4 < x < 6 ' + 4. Then -6 + 4 = y or 6 = §, or 6 + 4 = 5 or 6 = 1; thus 6=\. i| >x> i° or i° <x< 5. 22. Step 1: |x 2 - 3| < 0.1 => -0.1 < x 2 - 3 < 0.1 =4> 2.9 < x 2 < 3.1 => \J2.9 < x < a/3.1 Step 2: |x - 1/3 < 6 => -8 <x- y/3 < 8 => -6+y/3<x<6 + a/3. Then -6 + a/3 = a/2^9 =^> 6 = a/3 - a/Z9 w 0.0291, or 6 + a/3 = a/3T => 5 = thus 6 = 0.0286. 3.1- a/3 w 0.0286; Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley Section 2.3 Precise Definition of a Limit 81 23. Step 1: |x^ - 4| < 0.5 => -0.5 < x 2 - 4 < 0.5 ^ 3.5 < x 2 < 4.5 => V3-5 < |x| < V4.5 => - V4.5 < x < -^3.5 for x near —2. Step 2: |x - (-2)| <6 => -<5 < x + 2 < ,5 => -<5-2<x<<5-2. Then -6 - 2 = -y^ =^> 6 = ^5 - 2 « 0.1213, or 6 - 2 = thus<5 = a/45 -2« 0.12. '3.5 => <5 = 2- V3.5« 0.1292; 10 „„ 10 24. Stepl: |*-(-l)| < 0.1 => -0.1 < i + 1< 0.1 => - ±± < ± < - ^ =*► - ±2 > x > - f or - f < x in n • Step 2: |x-(-l)| <6 => -6 < x+ 1 < 6 =4- -S-Kx<6-1. Then -S - 1 = - f =>■ <5=i,or<5-l = -{f =4> § = i; thus <5 = i 25. Stepl: |(x 2 -5) - 11| < 1 => |x 2 - 16| < 1 =>■ -1< x 2 - 16 < 1 =^> 15 < x 2 < 17 =4- v^ < x < \/vi. Step 2: |x-4|<<5=^-<5<x-4<<5=>-<5 + 4<x<<5 + 4. Then -S + 4= v^ => <5 = 4 - v^ « 0.1270, or <5 + 4 = v^ =^ <5 = v^ - 4 w 0.1231; thus<5 = v^-4^0.12. 26. Stepl: |^-5| < 1 => -1 < ±f - 5 < 1 => 4< ±f <6 => ?>ifo>s => 30>x>20or20<x<30. Step 2: |x - 24 1 < 6 => -S < x - 24 < S => -6 + 24 < x < 6 + 24. Then -<5 + 24 = 20 => 6 = 4, or 5 + 24 = 30 => 6 = 6; thus ^6 = 4. 27. Step 1: |mx - 2m| < 0.03 =4> -0.03 < mx - 2m < 0.03 => -0.03 + 2m < mx < 0.03 + 2m =► 2 _OXB <x<2+ 0JB^ m m Step 2: \x-2\<6^-6<x-2<S^-S + 2<x<6 + 2. Then -<5 + 2 = 2 - ^ =>. ( 5 = PJ»^ + 2 = 2+ o : 03 ^ ^ = O03 ^ ith g = O03 m m ' m m ' m 28. Step 1: |mx - 3m| <c =>• -c<mx-3m<c => -c + 3m < mx < c + 3m => 3-^<x<3 + ^ Step 2: |x-3|<<5^-<5<x-3<<5^-<5 + 3<.r<<5 + 3. Then -6 + 3 = 3-^ =^ 5 = -^, or 6 + 3 = 3+^ => 6 = ±, In either case, 8 = ±. 1 m m " ' ' m m ' m 29. Step 1: |(mx + b) - (f + b) | < c => -c < mx - f < c =>• -c+f<mx<c+f =>• Step 2: |x-i|< ( 5=>-(5<x-i<5^>- l 5+i<x<6+i. 6=-,or<5+i = i + - =>- 5=-. In either case, 6 2 m ^ X ^ 2 + m- Then -<5 + \ = \ - *■ 2 2 m 30. Step 1: |(mx + b) - (m + b)| < 0.05 => -0.05 < mx - m < 0.05 =>• -0.05 + m < mx < 0.05 + m 1-^5 <x< 1 + ^. Step 2: \x—l\<6=>-6<x-l<6=>—6+l<x<6 + l. Then -6 + 1 = 1 - ^ => <5=^, r<S + l = l + 2:2§ =S> 6 = ^. In either case, 6 = 2^5. 31. lim (3 - 2x) = 3 - 2(3) x — ► 3 -0.02 < 6 - 2x < 0.02 =>• -6.02 < -2x < -5.98 =^ 3.01 > x > 2.99 or Stepl: |(3 -2x) - (-3)| < 0.02 : 2.99 < x < 3.01. Step 2: 0<|x-3|<(5 => -6 < x - 3 < 6 => -<5 + 3<x<<5 + 3. Then -6 + 3 = 2.99 =>■ <5 = 0.01, or S + 3 = 3.01 =^ <5 = 0.01; thus S = 0.01. 32. lim (-3x-2) = (-3)(-l)-2= 1 X — > — 1 Step 1: |(-3x - 2) - 1| < 0.03 => -0.03 < -3x - 3 < 0.03 => 0.01 > x + 1 > -0.01 => -1.01 < x < -0.99. Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesley 82 Chapter 2 Limits and Continuity Step 2: |x-(-l)|<<5 => -6<x+l<6 => -6-l<x<6-l. Then -6 - 1 = -1.01 => 6 = 0.01, or 6 - 1 = -0.99 => 5 = 0.01; thus S = 0.01. lim (x + 2) = 2 + 2 = 4, x ^ 2 x^ 2 33. lim — x^2 x " -4 -2 lim x^2 (x + 2)(x - 2) (x-2) Step 1: 1 V x *)• - 4 1 < 0.05 -0.05 < (x + 2)(x-2) (x-2) 4 < 0.05 =4> 3.95 < x + 2 < 4.05, x ^ 2 => 1.95 <x< 2.05, x^ 2. Step 2: |x - 2| < 6 => -S < x - 2 < <5 => -6 + 2 < x < 6 + 2. Then -8 + 2= 1.95 => <5 = 0.05, or 6 + 2 = 2.05 ^ 6 = 0.05; thus S = 0.05. 34. lim x^ -5 -6x + 5 lim (x + 5)(x+l) lim (x + 1) = -4, x / -5. -6x + 5 '\ i a\\ ^ r\ r\c _>. nn<^ (x + 5)(x + 1) Step 1: I ( x2 tf 5 +5 ) - (-4)1 < 0.05 => -0.05 < , y - -.,, .,,, .,._. (x ,, +4 < 0.05 =>• -4.05 < x+ 1 < -3.95, x/ -5 => -5.05 <x < -4.95, x^ -5. Step 2: |x - (-5)| <6 => -S <\ + 5 <6 =4> -6-5<x<6-5. Then -6 - 5 = -5.05 ^ 6 = 0.05, or <5 - 5 = -4.95 => <5 = 0.05; thus <5 = 0.05. 35. lim Vl -5x = y/l - 5(-3) = v^ = 4 x^ -3 Step 1: I a/I -5x - 4| < 0.5 => -0.5 < \/\ - 5x - 4 < 0.5 =» 3.5 < v 7 ! - 5x < 4.5 => 11.25 < -5x < 19.25 => -3.85 < x < -2.25. Step 2: |x - (-3)| <6 => -S < x + 3 <6 => -<5-3<x<<5-3. Then -6-3 = -3.85 => 6 = 0.85, or 6 - 3 = -2.25 => 0.75; thus 6 = 0.75. 12.25 < 1 - 5x < 20.25 10 16 W 24 -^>x>12or|<x<§. 36. lim 4 - = i = 2 x^2 x 2 Step 1: || -2| <0.4 => -0.4 < | Step 2: |x-2|<<5^-<5<x-2<<5^.-<5 + 2<x<<5 + 2. Then -6 + 2 = f => <5 = |, or 6 + 2 = | => 6 = | ; thus «5 = 5 ■ 37. Step 1: |(9-x)-5|<e=>-e<4-x<6=>-e-4<-x<e-4=>e + 4>x>4-e=>4-e<x<4 + e. Step 2: |x-4|<<5=^-<5<x-4<<5=^-<5 + 4<x<<5 + 4. Then -6 + 4 = -e + 4 =^ <5 = e, or<54-4 = e + 4 =>■ <5 = e. Thus choose <5 = e. 38. Step 1: |(3x — 7) — 2| < e =>• -e < 3x - 9 < e =>■ 9 - e < 3x < 9 + e => 3-f<x<3 + f. Step 2: |x-3|<<5^-<5<x-3<<5^-<5 + 3<x<<5 + 3. Then -6 + 3 = 3 - f => <5=§,or<5 + 3 = 3+§ =>• § = f . Thus choose <5 = f . 39. Step 1: k/x - 5 - 2 < e => -e < \fk - 5 - 2 < e => 2-e < a/x - 5 < 2 + e => (2 - e) 2 < x - 5 < (2 + e) 2 =4> (2 - e) 2 + 5 < x < (2 + e) 2 + 5. Step 2: |x-9|<<5=^-6<x-9<<S=^-£ + 9<x<<S Then -<5 + 9 = e 2 - 4e + 9 ^ 5 = 4e - e 2 , or <5 + 9 the smaller distance, S = Ae — e 2 . e 2 + 4e 4- 9 =$> 6 = 4e + e 2 . Thus choose 40. Step 1: \j A - x - 2 < e => -e < ^4 - x - 2 < e => 2 - e < ^4 - x < 2 + e =^ (2 - e) 2 < 4 - x < (2 + e) 2 => -(2 + e) 2 < x - 4 < -(2 - e) 2 => -(2 + e) 2 + 4 < x < -(2 - e) 2 + 4. Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley Section 2.3 Precise Definition of a Limit 83 Step 2: |x-0| <6 => -8 < x < 8. Then -S = -(2 + e) 2 + 4 = -e 2 - 4e =4> <5 = 4e + e 2 , or 5 = -(2 - e) 2 + 4 = 4e - e 2 . Thus choose the smaller distance, 8 = 4e — e 2 . H. Step 1: Forx^ 1, |x 2 - 1| < e => -6 < x 2 - 1 < e => 1 - e < x 2 < 1 + e => v/l ~ e < l x l < V 1 + e =>- V 1 — e < x < v 1 + e near x = 1. Step 2: |x - 1| < <5 =>- -<5 < x - 1 < <5 => -<5 + 1 < x < 6 + 1. Then -6 + 1 = sjl -e =>• 8 = 1- \fl - e, or 8 + 1 = y/l + e => 8 = yjl + e - 1. Choose 8 = min < 1 — \J 1 — e, yl+e — 1 \ , that is, the smaller of the two distances. 42. Step 1: For x ^ -2, |x 2 - 4| < e =>■ -e < x 2 - 4 < e => 4 - 6 < x 2 < 4 + e =^> ^4 - e < |x| < a/4 + e ^> -i/4 + e<x<- a/4~ e near x = —2. Step 2: |x - (-2)| <S =>• -8 <x + 2<8 =>• -<5-2<x<<5-2. Then -6 - 2 = -a/4 + e =>• 6 = y/Z + e - 2, or <5 - 2 = -a/4- e => 5 = 2- a/4 - e. Choose <S = min { a/4+7 - 2, 2 - a/4 - e} . l-l|<e => -6 <i-Kc => l- e <i<l + e ^ T i 7 <x< T + 7 . 43. Step 1: Step 2: |x-l|<<5=>-<5<x-l<<5^1-<5<x<l+<5. Thenl-^^ =» (5 = l- T i 7 = T ^,orl + 5 = T i 7 Choose (5 = y+^< me smaller of the two distances. 6= 1-^-1= r 5 -- 1 — e 1 — e 44. Step 1: 1 _| ^- ,. -_> c <^ i i^-f^ 1 f ^ x ^ x 4- f __* l-3c ^ 1 ^ 1+3- . 3 . 2 ^ 3 -J--|<£ _^ _ £< _ J __ <e ^ __ £< _ j< _ +£ =^ _____ < __ < _____ ^ _____ > X > j^gj TT37 < l x l < VT^' or TT37 <X< ^t^ for x near ^3. Step 2: |x- a/3| <5 =» -6<x- v / 3<6 => v / 3-<^<x<V / 3 + <5. Thenv^-« = VTT5I =► * = <fi " VOT or V^ + « = V T^C => « = V T^3_ " V^ Choose 8 = min -^ a/3 - 3 / 3 1 + 3f ' V 1 - 3e 3 . 45. Step 1: I (^f ) - (-6)1 <e => -e < (x - 3) + 6 < e, x ^ -3 =4> -e < x + 3 < e Step 2: |x - (-3)| <8 => -8 < x + 3 <8 => -<5 - 3 < x < <5 - 3. Then — 8 — 3 = — e — 3 => 8 = e, or 8 — 3 = e — 3 => 8 = e. Choose 8 = e. 46. Step 1: I (^ff) - 2| < e =>■ -e < (x + 1) - 2 < e, x ^ 1 =4> l-e<x<l + e. Step 2: |x-l|<<5=>-<5<x-l<<5=^l-<5<x<l+<5. Then 1 — 8 = 1 — e => 8 = e, or 1 + 6 = 1 + e =$■ 8 = e. Choose 8 = e. => -e-3<x<e-3. 47. Step 1: x < 1: |(4 - 2x) - 2| < e => < 2 - 2x < e since x < 1. Thus, 1 - f < x < 0; x > 1: |(6x - 4) - 2| < e ^> 0<6x-6<e since x > 1. Thus, 1 < x < 1 + |. Step 2: |x-l|<<5^>-(5<x-l<<5=^l-6<x<l+<5. Then 1 — ^ = 1 — § ^> 5 = f , orl + <S=l + f ^ 5 = f . Choose 8 = f . 48. Step 1: x < 0: |2x - 0| < e => -e < 2x < ^> - § < x < 0; x >0: < e => <x< 2e. Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley 84 Chapter 2 Limits and Continuity Step 2: |x-0| <6 => -6 < x < 6. Then —6 | =>. 5 = §, or <5 = 2e => 5 = 2e. Choose S = § . 49. By the figure, — x < x sin 1 < x for all x > and — x > x sin - > x for x < 0. Since lim (— x) = lim x = 0, x x x^Ox^O then by the sandwich theorem, in either case, lim x sin - = 0. J x^ x 50. By the figure, — x 2 < x 2 sin - < x 2 for all x except possibly at x = 0. Since lim (— x 2 ) = lim x 2 = 0, then by the sandwich theorem, lim x 2 sin - = 0. J x^O x 51. As x approaches the value 0, the values of g(x) approach k. Thus for every number e > 0, there exists a S > such that < |x - 0| < 6 => |g(x) - k| < e. 52. Write x = h + c. Then 0<|x-c|<<5<^>-<5<x-c<<5, x ^ c <J4> -<5 < (h + c) - c < <5, h + c^c ■^ -S < h < S, h / ^ < |h - 0| < 6. Thus, limf(x) = L •£> for any e > 0, there exists 6 > such that |f(x) — L| < e whenever < |x — c| < 6 ^lf(h ■ c) - Ll < e whenever < |h - 0| < 6 ^> limf(h + c) = L. h^0 53. Letf(x) = x 2 . The function values do get closer to — 1 as x approaches 0, but lim f(x) = 0, not — 1 . The x — > function f(x) = x 2 never gets arbitrarily close to — 1 for x near 0. 54. Let f(x) = sin x, L = |, and Xo = 0. There exists a value of x (namely, x = |) for which | sin x — 1 1 < e for any given e > 0. However, lim sin x = 0, not |. The wrong statement does not require x to be arbitrarily close to x — > z Xo. As another example, let g(x) = sin j, L = I, and Xo = 0. We can choose infinitely many values of x near such that sin - = \ as you can see from the accompanying figure. However, lim sin - fails to exist. The x i x — > x wrong statement does not require all values of x arbitrarily close to x = to lie within e > of L = | . Again you can see from the figure that there are also infinitely many values of x near such that sin - = 0. If we choose e < \ we cannot satisfy the inequality | sin 1 — \ | < e for all values of x sufficiently near Xo = 0. y 55. A -91 <0.01 -0.01 < 7Ti -9 < 0.01 8.99 < ^f < 9.01 ^(8.99)<x 2 < 4(9.01) =$■ 2a /^ < x < 2i i^r or 3.384 < x < 3.387. To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 56. V = RI =>• ¥ =1 =>■ || -5| <0.1 => K IK I — -0.1 < 120 5 <0.1 4.9 < l -f <5.1 49 R 120 51 Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesley Section 2.3 Precise Definition of a Limit 85 (120X10) < R < (120X10) 23.53 < R < 24.48. 51 - " - 49 To be safe, the left endpoint was rounded up and the right endpoint was rounded down. x. Then |f(x) - 2| 2-x>2-l = l. That is, x > 1. That is, 57. (a) -6 < x - 1 < => 1 - 6 < x < 1 => f(x) jf(x) — 2 1 > 1 > i no matter how small S is ta (b) < x - 1 < 8 => 1 < x < 1 + 6 => f(x) = |f(x) — 1 1 > 1 no matter how small S is taken when 1 < x < 1 + <5 = (c) -6 < x - 1 < => 1 - 8 < x < 1 =>• f(x) = x. Then |f(x) - 1.5| = |x - 1.5| = 1.5 - x > 1.5 - 1 = 0.5. Also, 0<x-l<<5 => 1<x<1 + <5 => f(x) = x+l. Then |f(x) - 1.5| = |(x + 1) - 1.5| = |x - 0.5| = x — 0.5 > 1 — 0.5 = 0.5. Thus, no matter how small S is taken, there exists a value of x such that -6 < x - 1 < S but |f(x) - 1.51 > | =>• lim f(x) / 1.5. 1 X — > 1 ken when 1 - S < x < 1 =>■ lim f(x) ^ 2. X — > 1 x+1. Then|f(x)-l| = |(x+l)-l| = |x| lim f(x)/l. X — > 1 |h(x) - 4| = 2. Thus for e < 2, |h(x) - 4| > e whenever 2<x<2 + <5no => lim h(x) ^ 4. x^ 2 ^ |h(x) - 3| = 1. Thus for e < 1, |h(x) - 3| > e whenever 2 < x < 2 + 8 no » lim h(x) ^ 3. x^ 2 ^ (c) For 2 - <5 < x < 2 =*• h(x) = x 2 so |h(x) - 2 1 - u - when x is near 2 and to the left on the real line =^ |x 58. (a) For 2 < x < 2 + S =$> h(x) = 2 =; matter how small we choose 6 > (b) For 2 < x < 2 + 6 => h(x) = 2 =1 matter how small we choose 6 > \- — 2 1 . No matter how small S > is chosen, x 2 is close to 4 2 2| will be close to 2. Thus if e < 1, |h(x) - 2| > e whenever 2 — <5<x<2no mater how small we choose (5 > lim h(x) ^ 2. x^2 ^ 59. (a) For 3 - S < x < 3 =^> f(x) > 4.8 => |f(x) - 4| > 0.8. Thus for e < 0.8, |f(x) - 4| > e whenever 3 — <5<x<3no matter how small we choose 6 > lim f(x) / 4. x — > 3 (b) For 3 < x < 3 f(x) < 3 => |f(x) - 4.8| > 1.8. Thus for e < 1.8, |f(x) - 4.8| > e whenever 3 < x < 3 + S no matter how small we choose 6 > lim f(x) + 4.8. x — » 3 (c) For 3 - 6 < x < 3 => f(x) > 4.8 => |f(x) - 3| > 1.8. Again, for e < 1.8, |f(x) - 3| > e whenever 3 - 6 < x < 3 no matter how small we choose 6 > lim f(x) / 3. x — » 3 60. (a) No matter how small we choose 6 > 0, for x near — 1 satisfying — 1 — 6 < x < — 1 + 6, the values of g(x) are near 1 => |g(x) — 2| is near 1. Then, for e = \ we have |g(x) — 2| > | for some x satisfying -1 -6 < x < -1 + <5,or0 < |x+ 11 < S => lim g(x) ^ 2. X — > — 1 (b) Yes, lim g(x) = 1 because from the graph we can find a 6 > such that |g(x) — 1 1 < e if < |x — (— 1)| < 8, 61-66. Example CAS commands (values of del may vary for a specified eps): Maple : f := x -> (x A 4-81)/(x-3);x0 := 3; plot( f (x), x=x0- 1 . .x0+ 1 , color=black, # (a) title="Section 2.3, #6 1(a)" ); L := limit( f(x), x=x0 ); # (b) epsilon := 0.2; # (c) plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01, color=black, linestyle=[l,3,3], title="Section 2.3, #6 1(c)" ); q := fsolve( abs( f(x)-L ) = epsilon, x=x0-l..x0+l ); # (d) delta := abs(xO-q); plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e) Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesley 86 Chapter 2 Limits and Continuity q := fsolve( abs( f(x)-L ) = eps, x=xO-l..xO+l ); delta := abs(xO-q); head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta ); print(plot( [f(x),L-eps,L+eps], x=xO-delta..xO+delta, color=black, linestyle=[l,3,3], title=head )); end do: Mathematica (assigned function and values for xO, eps and del may vary): Clear[f, x] yl: = L eps; y2: = L + eps; xO = 1; f[x_]: = (3x 2 - (7x + l)Sqrt[x] + 5)/(x - 1) Plot[f[x], {x, xO - 0.2, xO + 0.2}] L: = Limit [f[x], x -> xO] eps = 0.1; del = 0.2; Plot[{f[x], yl, y2},{x, xO - del, xO + del},PlotRange ->• {L - 2eps, L + 2eps}] 2.4 ONE-SIDED LIMITS AND LIMITS AT INFINITY 1 . (a) True (e) True (i) False (b) True (f) True (j) False (c) False (g) False (k) True (d) True (h) False (1) False (a) True (e) True (i) True (b) False (f) True (j) False (c) False (g) True (k) True (d) True (h) True (a) lim f(x) = | + 1 = 2, lim f(x) = 3-2 = 1 x -» 2+ 2 x -> 2- (b) No, lim f(x) does not exist because lim f(x) ^ lim f(x) x ^ 2 W x ^ 2+ x ^ 2" W (c) lim f(x) = % + 1 = 3, lim f(x) = £ + 1 = 3 x -> 4" - x -> 4+ £ (d) Yes, lim f(x) = 3 because 3 = lim f(x) = lim f(x) x ^ 4 x^4~ x ^ 4+ (a) lim f(x) = | = 1, lim f(x) = 3 - 2 = 1, f(2) = 2 x -> 2+ l x -> 2- (b) Yes, lim f(x) = 1 because 1 = lim f(x) = lim f(x) x ^ 2 x ^ 2+ x ^ 2 (c) lim f( x ) = 3-(-l) = 4, lim f(x) = 3 - (-1) = 4 x — > -1 X — > -1 + (d) Yes, lim f(x) = 4 because 4 = lim f(x) = lim f(x) X— > -1 X -> -1 X -» -1+ 5. (a) No, lim f(x) does not exist since sin Q) does not approach any single value as x approaches (b) lim f(x) x — > lim x->0- (c) lim f(x) does not exist because lim f(x) does not exist x -> x -» 0+ 6. (a) Yes, lim g(x) = by the sandwich theorem since — */x < g(x) < ^/x when x > x — > 0+ (b) No, lim _ g(x) does not exist since \/x is not defined for x < x — > (c) No, lim g(x) does not exist since lim g(x) does not exist x — > x — > Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesley Section 2.4 One-Sided Limits and Limits at Infinity 87 7. (a) 0, x=\ (b) lim f(x) X — ► 1 1 lim f(x) : — > 1+ (c) Yes, lim f(x) x — > 1 1 since the right-hand and left-hand limits exist and equal 1 8. (a) (b) lim f(x) = X — > 1 + lim f(x) (c) Yes, lim f(x) = since the right-hand and left-hand X — > 1 limits exist and equal (a) domain: < x < 2 range: < y < 1 and y = 2 (b) lim f(x) exists for c belonging to (0,1)U(1,2) (c) x = 2 (d) x = 3 f 2 y = it-x 2 , 0£x<l 1, l<;t<2 2, x = 2 • "N \> • > 1 2 10. (a) domain: — oo < x < oo range: — 1 < y < 1 (b) lim f(x) exists for c belonging to (-oo, -l)U(-l,l)U(l,oo) (c) none (d) none x, -1 <^x < or < x <_1 ^\ x = 0, x < 1 or x > -1 11. lim x^ -0.5" x + 2 x- 1 -0.5 + 2 -0.5 + 1 m = \/3 1/2 V J 12. lim J±=± x^i+ V x + 2 l- l 1 + 2 ^=0 13- x jim 2+ (^r) (m = (=§£l) {Mjk) = <3 (I) 14. l»n- (l+T) (^) (^) = (ih) W) W) = (*) (D (?) 15. lim h->0+ Vh 2 + 4h + 5 - V5 lim h-»0+ ( y/h 2 + 4h + 5-y/E\ /y/h 2 + 4h + 5 + y/5 \ / y/h 2 + 4h + 5+y/5 \ / V Vh 2 + 4h + 5+x/5/ lim (h 2 + 4h + 5)-5 lim h(h + 4) + 4 0+ h(Vh 2 +4h + 5+\/5) h-t0+ h(Vh 2 + 4h + 5 + v/5) 75 + V^ V^ Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley 88 Chapter 2 Limits and Continuity 16. lim h->0" lim v /6- x /5h2+llh + 6 _ ,. / y/e-x/Sh^+llh + e X / v/6+v/5h2+llh + 6 \ lf '" V h ) \j6+s/S&+Uh + 6) -h(5h + U) _ -(0+11) _ h->0 6-(5h 2 +llh + 6) lim I) IF h I /6+\/5h2+llh + 6) h-»0 _ h(x/6 + V5h 2 + llh + 6) \/6+ v^ W6 (|x + 2| = x + 2forx > -2) 17. (a) Hm (x + 3) ^ = Km (x + 3) §±| -2+ -2+ lim (x + 3) = (-2) + 3=1 x^ -2+ (b) x jim 2 _(x + 3)^ = Um_(x + 3)[=£ -(x + 2) 2) (|x + 2| = -(x + 2)forx < -2) lim (x + 3)(-l) = -(-2 + 3) = -l x — > — 2 18. (a) lim /2x(x-l) lim y/2x~(x-l) X-T1+ I-- !| x"l+ < X -D = lim v2x = v2 x^ 1+ (b) lim 4^^ = lim ^ {x ~n |x - 11 = x - 1 forx > 1) |x- 1| = -(x- l)forx< 1) lim — V 2x = — v 2 x — > 1 19. (a) lim HI - 3 3 (b) e lim_ f = l 20. (a) lim (t - Itl) =4-4 = t -> 4+ V L J ^ (b) t lim_(t-LtJ) = 4-3=1 21. lim ^^ = lim ™* = 1 (where x = y^tf) 6(^0 V2B x^O x V V7 22. lim SS^J = lim ^^ t -» ' t -» o kt lim (9^0 ^ = k lim ^ 23. lim ^ = i lim ^^ = ? lim ^ = | lim k- 1 =k sinfl _ 3 ■ o 4 y ■ o 3 y o 3 y 4 e -> o (where 9 = kt) (where 9 = 3y) 24. lim h-»0" sin 3h lim h-»0" 'I . .J*-) - I ii m i ,3 si „3hJ " 3 h ™ - (-») lim (9-0- 1 i (where (9 = 3h) 25. lim Sa^ = hm liSS) = hm _Ms^_ = ( i im i ) ( i im 2_sm2x\ =1 . 2 = 2 x^0 x x^0 x x^O xcos2x Vx^O co * 2x J Vx^O 2x / 26. lim A = 2 lim -^ t^O tant t^Q (Mil) 2 lim ^J = 2 ( lim cos ( t -> Sln ' Vt -> )W 2-1-1 = 2 27. lim ^^ = lim (-V x ¥ n cos 5x x ^ n V sin 2x cos i ) = (l i im * ") ( Hm i ) = (1-1) (1) = 1 3 5x / V 2 x > q sin 2x y V x _;, q cos 5x / V 2 / v / 2 28. lim 6x 2 (cot x)(csc 2x) = lim 6x2cosx = lim (3 cos x • -^ ^-) =3-1-1=3 _ r, ■■ „ n Sin X sm 7x a V sin X Sin ZX / x^0 x j, q sin x sin 2x x y n 29. lim x + xcosx = lim (-^-* — + . xc ° sx = km (J L_) + hm -J- x ^ n sin x cos x x ^ n \ sin x cos x sin x cos x / x ^ n V sin x cos x / x y n sin x lim ' ! x^0 x li- (^)+ x ^ ( + )=(1)(D+ 1=2 30. lim x^0 2x lim , x^0 V2 I + I(fc))=0-1 + !<1) = Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley Section 2.4 One-Sided Limits and Limits at Infinity 89 31. lim t->o sin(l — cos t) 1— cos t lim 8^0 1 since 9=1— cos t — > as t — > 32. lim h->0 sin (sin h) sin h lim 0^0 1 since 9 = sin h — » as h — > 33. lim 42|j = lim sin 9 20\ _ 1 i- /sinfl 20 \ _ 1 i i _ 1 sin 20 fl""*n V sin 20 '2d) 2 o J} n V sin 20 ) 2 ' i ' l 2 6*^0 34. lim ^ = lim (4^ • ^ • z) x j, n sin 4x x ► sin 5 U (sin5x . _4x > 4 y ► Q ^ ^x sin 4x / 5.1-1=5 4 A J- 4 35. lim ^ = lim (^ - ^V) = lim (^ . i . *E . |> - sin fix Y , a V cos 3x sin 8x / Y , a V cos 3x sin 8x 3x 8 1 x^O lim x^O 1 \ / sin 3x \ / 8x x^O \ cos 3x 7 V 3x 7 V sin 8x7 1-1-1 36. lim sm3y "f 5y = lim sin 3y si " 4y c ° s 5y = lim ( s -^ y _> y cot 4y y _, g y cos 4y sin 5y y _, g ^ y \ ( sin 4y \ / cos5y \ / 3-4-5y \ ^ ^cos4 y/ / \^sin5y,/ ^3-4-Sy^ y lim o (^)(^)( i || y )(^|)(M) = l. 1 .l. 1 .n = n Note: In these exercises we use the result lim -ij = whenever m > 0. This result follows immediately from v _^ 4- m x"V" n J ± oo x Example 6 and the power rule in Theorem 8: lim ( -4 F F X -» ± 00 v x lim (I) m/n =( lim iV X — > ± 00 X VX — > ± CO X / m/„ = Q_ 37. (a) -3 (b) -3 38. (a) 7T (b) 7T 39. (a) i 40. (a) 1 (b) § 41. (a) -§ (b) 42. (a) | (b) 1 43. - i < 1 ^ sin 2x ^ 1 sin 2x lim X — > 00 x by the Sandwich Theorem 44. 1 £>■ COS ^ 1 30 — 30 — 30 lim e^ -c 30 by the Sandwich Theorem 45. lim 2 r, t + si '" = lim -1+m t^oo t + cost t-»00 l + C^f 1 ) 0-1+0 _ _i 1+0 ' 46. lim , ! + sh ". = lim - V ( !vL r — > oo 2r + 7-5sinr r _ > oc 2+^-5(^ li m i+o = I r l -H% 2 + 0-0 2 47. (a) lim ^±| = lim |±| = f (b) | (same process as part (a)) 48. (a) lim 2x 3 + 7 lim *oo x 3 — x 2 + x + 7 (b) 2 (same process as part (a)) 00 !-J + 7i + 73 Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesley 90 Chapter 2 Limits and Continuity (a) lim 4±\ ..... , v/ x ^ oo x 2 + 3 x ^ oo 1 + 4 1 , i lim_ ff£ = 3 (b) (same process as part (a)) 50. (a) lim ^-^ = lim f^ x/ X — > OO x z - 2 x — > 00 1 — - (b) (same process as part (a)) 51. (a) lim 7x 3 x — > oo x 3 - 3x 2 + 6x x — > oo 1 - I + lim 7 (b) 7 (same process as part (a)) 52. (a) lim x — > oo x 3 - 4x + 1 x lim 1 _ 4 , 1 — U oo 1-^ + (b) (same process as part (a)) 53. (a) lim 10x5 + f + 31 = lim 1±A±1 = «) X — > oo x u X — > oc 1 (b) (same process as part (a)) 9x 4 + > 9 + 54. (a) lim , 4 ^T 2 +X ,,- = lim T —, , — - - - v/ x — > oo 2x 4 + 5x 2 - x + 6 x ^ oo 2 + ^-4j + -^ 2 (b) | (same process as part (a)) 55. (a) lim ~ 2 /'rf + 3 = lim \ ? + fi = - \ w x-»x 3x J + 3x- ! -5i x — > oo 3 + - — ?j 3 (b) — § (same process as part (a)) 56. (a) lim —, — _ ., x _ , , « = lim - — , — k — g- = — 1 v ' x — > oo x 4 - 7x 3 + 7x 2 + 9 x — > oo 1 - z (b) —1 (same process as part (a)) -l 57. lim 2,/x + x- 1 lim x — > oo 3x-7 x — > oo 3 58. lim lim x — » oo 2—Jx x — > oo ( 2 , i i v \7m) 59. lim yx--x/x- lim l_ x (l/5)-(l/3) lim v-. 2 v - 00 ^/x+^/x X — > -00 1 +x(V5)-(l/3) x _» _oc j j t 1 \ 60. lim lim X^OCX 2 -x 3 x ^ oo 1 oo 6 1. lim 2xV 3 -xV 3 + 7 =Jim 2xV15 -^ + ^ x ™oo x 8 / 5 + 3x + y/x x — >"bo 1 + ^E + ^S~j OO ,-n, ,- ?/x-5x + 3 .. 72/3 ~ 5 +x 62. lim , v , 2/3 -, = lim f^ T x — > — oo 2x + x 2 / 3 -4 x — > — oo 2+-TH-- 63. Yes. If lim f(x) = L = lim_ f(x), then lim f(x) = L. If lim f(x) ^ lim_ f(x), then lim f(x) does not exist. x » a~t~ x — > a x > a ^ ^ g+ x — > a. x > a. 64. Since lim f(x) = L if and only if lim f(x) = L and lim_ f(x) = L, then lim f(x) can be found by calculating X > C Y i f*T X — > C X ► C lim f(x). 65. If f is an odd function of x, then f(— x) = — f(x). Given lim f(x) = 3, then lim f(x) = — 3. x -> 0+ x -» 0- ' Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley Section 2.4 One-Sided Limits and Limits at Infinity 91 66. If f is an even function of x, then f(— x) = f(x). Given lim f(x) = 7 then lim f(x) = 7. However, nothing x -> 2- x -> -2+ '6 can be said about lim f(x) because we don't know lim f(x). x -> -2" x -> 2+ 67. Yes. If lim ^ X — » oo g(x) 2 then the ratio of the polynomials' leading coefficients is 2, so Jim 2gr 2 as well. 68. Yes, it can have a horizontal or oblique asymptote. 69. At most 1 horizontal asymptote: If lim f(x) x — > i 6o g(x) L, then the ratio of the polynomials' leading coefficients is L, so lim fix) x — > -oo g(x) L as well. 70. lim vx 2 + x — v x 2 — x = lim i/x 2 + x — y x 2 — x • X ^ OO v v x — > OO [ v \/x 2 + x + \/x 2 — X lim (x 2 + x) - (x 2 - x) \/x 2 + x + \/x 2 - x J X — > 00 ^x 2 + x + v/x 2 lim 2x V x 2 + x 4- \/x 2 — x X lim OO ./ill l + 7+\/l- 1 ~~ 1 + 1 71. For any e > 0, take N = 1. Then for all x > N we have that |f(x) - k| = |k - k| = < e. 72. For any e > 0, take N = 1. Then for all y < -N we have that |f(x) - k| = |k - k| = < e. 73. I = (5, 5 + 6) =4> 5 < x < 5 + 6. Also, ^\ - 5 < e => x - 5 < e 2 =4> x < 5 + e 2 . Choose <S = e 2 => lim \/ x ~ 5 = 0. x^5+ 74. I = (4 - 6, 4) =>• 4 - 6 < x < 4. Also, ^/4 - x < e =>• 4 - x < e 2 =4> x > 4 - e 2 . Choose 6 = e 2 => lim_ \/4 - x = 0. x — > 4 75. As x — > _ the number x is always negative. Thus, A — (— 1) < e =>• | ^ + l| < e =^> 0<e which is always true independent of the value of x. Hence we can choose any S > with — 6 < x < =>■ lim A = — 1. x — > l x l 76. Since x — > 2 + we have x > 2 and |x — 21 = x — 2. Then, x-2 |x-2| - 1 x-2 x-2 1 <e=> 0< e which is always true so long as x > 2. Hence we can choose any 6 > 0, and thus 2 < x < 2 + 5 x-2 x— 21 1 < e. Thus, lim A=| = 1 x^ -2 + |x-2| 77. (a) lim [ X J = 400. Just observe that if 400 < x < 401, then [xj = 400. Thus if we choose 6 = 1, we have for any x — > 400 + number e > that 400 < x < 400 + 6 => \ |xj - 400| = |400 - 400| = < e. (b) lim I x I = 399. Just observe that if 399 < x < 400 then |x|= 399. Thus if we choose S = 1, we have for any number e > that 400 - 6 < x < 400 => | LxJ - 399| = |399 - 399| = < e. (c) Since lim I x I ^ lim I x I we conclude that lim I x I does not exist. x _> 400+ L J x -» 400" L J x -> 400 L J 78. (a) lim f(x) = lim ^/x = \/0 = 0; | y^x — 0| < e =^ -e < y/x < e => < x < e 2 for x positive. Choose 6 = e 2 => lim f(x) = 0. x^0+ (b) lim f(x) = lim x 2 sin (-) = by the sandwich theorem since — x 2 < x 2 sin (-) < x 2 for all x ^ 0. x^0 x^0 vx/ • vx/ Since |x 2 — 0| = j— x 2 — 0| = x 2 < e whenever |x| < yj~e, we choose 6 = yj~e and obtain |x 2 sin (ij — 0| < e if -6 < x < 0. Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesley 92 Chapter 2 Limits and Continuity (c) The function f has limit at Xq = since both the right-hand and left-hand limits exist and equal 0. 79. lim xsini=lim \ sin 8 = 1, (8 = ±) 80. lim ^4= lim f 5 ^ = : 1, 0^0 -oo 1 + 1 g "q- ! + e ! i) 81. lim X — > ± OO 3x + 4 2x-5 lim x — > ± 00 3 + ! lim |±| = | t-»0 2 ~ 5t 2 (t=D 82. lim ( i ) lA = lim z z = 1, (z = ±) J^M «' z ^ 0+ X 83. lim (3 + |) (cos i) = lim (3 + 20)(cos 0) = (3)(1) = 3, (0 = 1) x^ ±00 x x 6* — » 84. lim (4 -cos i) (l + sin 1 ) = lim (30 2 - cos 0) (1 + sin 0) = (0 - 1)(1 + 0) = -1, X — > OC ^ x X / V X / r\ q-[- 2.5 INFINITE LIMITS AND VERTICAL ASYMPTOTES 1. lim x^0+ OO (positive \ positive J 2. lim x^O" 2x -DC: positive \ negative J 3. lim x^2" jc-2 -oo (positive \ negative I 4. lim — -; = oo x -> 3+ X ~ J ( positive \ positive y 5. lim x^ -8 2x + x+8 -'00' (negative \ positive J 6 - x Jj m 5 - 2x+T0 = °° (negative \ negative J 1. lim , \. 2 = oo 9. (a) lim x - 0+ 2 3xV3 - oo 10. (a) lim x -> 0+ 2 xV= - oo 11. lim -Ar = x -» x ' = lim x^O 4 (xV5f 13. lim tan x = oo (positive \ positive J 8. lim x ^0 x '( x + 1 ) (b) lim Q- 3xV3 (b) lim : -oo x^O" xV5 12. lim -L = lim —I-, x^O x2/3 x^O (xV 3 ) 2 14. lim sec x = oo (negative \ positive-positive J 15. lim (1 + csc 6) = — oo 16. lim (2 — cot 0) = — oo and lim (2 — cot 8) = oo, so the limit does not exist e -> o+ e^o- 17. (a) x hm + x i L 4= x lim + ^^j (b) lim ^ = lim (x+2 x x * = ^ x _» 2" (x+2)(x-2) 1 (C) x i! m 2+ ^4= x _L im 2+ (x + 2)(x-2) (d) lim a lim 2" * 2 -4 x -T-2 - (*+2)(x-2) f .. 1 .. ) \ positive-positive J ( .. i ) V positive -negative / f .. i . ) ^ positive-negative y r . i . 1 V negative -negative / Copyright (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesley Section 2.5 Infinite Limits and Vertical Asymptotes 93 18. (a) lim lim x ""+ * 2 -l x "" + ( X + 1 K"-D oo (b) lim Jtx X — > 1 x ' lim , , * — rv x _» 1~ (x+l)(x-l) = lim -oo (c) lim -fa - ,„„ x^-l + x ' x— >-l + ( X+1 X X -!) (d) lim x^ -1- X 2 -l lim _ x - (x+l)(x-l) -OO 19. (a) lim f - i = x^0+ 2 (b) lim 4 ~ l = x -> 0" 2 x (c) lim £ - i s /^ 2 x x^ yi (d) lim £ - i = x--l 2 20. (a) lim x2 -! - X TV 2x + 4 (c) lim x2 -! = x" 1+ 2x + 4 (d) lim " 2_1 = xTo- 2x + 4 21. (a) lim + 4^W x _> 0+ x 2x (b) lim x V x + 2 2 x _> 2+ x 2x (c) lim x2 3 " 3x , + 2 2 x -> 2- x 2x (d) lim x2 -3 x + 2 . n X 3 - 2x 2 " x — > 2 x zx lim x^ 4 lim x^ 0- 1 2- 3 2 1 2 l/3 -oo ■ = oo 2-V3 _ 2-1/3 positive \ positive-positive J positive \ positive- negative J negative \ positive -negative negative negative -negative ative / M ltive y positive (i) 3 lim (x+l)(x-l) 2x + 4 positive \ positive y 2-0 2 + 4 (b) ijm 2 _ f^ = -oo (e) lim o x 2 - 3x + 2 x 3 — 2x 2 lim .-»0 (x-2)(x-l) f x 2 (x-2) (x-2)(x-l) x ^ 2 + x2 ( x " 2 > = lim x^2 lim x^2 lim (x-2)(x-l) x 2 (x - 2) (x-2)(x-l) _ x 2 (x - 2) (x-2)(x-l) x 2 (x - 2) = — oo / negative-negative \ positive-negative = lim ^- = i x ^ 2 x^2+ x2 4 ^ = lim i^i = i x ^ 2 x^2- " 2 4 ' r lim M = 1 x ^ 2 x -» 2 x2 4 ' r — OO / negative-negative f positive \ I negative J x 2 - 3x + 2 22 - ( a > lim 9+ ^^4x- (b) (c) (d) (c) 2 lim x 2 - 3x + 2 lim x^0 o+ x 3 — 4x 2 -3x + 2 x 3 -4x lim (x - 2)(x - 1) = ._T2+ x ( x -2)( x + 2) lim < X - 2 >( X -D x _^_2+ x ( x " 2 X x + 2) lim (x-l) lim x 2 - 3x + 2 (x-2)(x-l) 5- x(x-2)(x + 2) lim fr- 2 **- 1 ) x Tl+ x ( x "2)(x + 2) lim .— x^l+ x3 - 4x lim x ~ \, = —00 x-0+ x(x + 2) and lim x ~ \, = 00 x — > _ x ( x + 2) so the function has no limit as x ^2+ x(x + 2) 2(4) lim (x x^-2+ x(x -1) _ + 2) im ->tr (x-l) x(x + 2) = OO . (x-l) -»1+ x(x + 2) (1)(3) OO 0. (negative \ positive-positive J (negative negative-positive I negative \ I negative- positive J I negative \ I negative -positive J 23. (a) lim \l t -> 0+ L -4-1 tV 3 J (b) lim \2 4-1 tV 3 J 00 24. (a) lim [4j+7] t-»a (b) t lim_ [^+7] 25. (a) lim [4 x ^o+ L x/ (c) lim [-± x ^i+ L x2/ (x-l) 2 ' 3 2 x 2 /- 3 ' (x - 1) 2 / 3 (d) x^l" [^ (x- l) 2 /' 3 2 x 2/3 T- ( X _ 1)2/3 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 94 Chapter 2 Limits and Continuity 26. (a) lim [+, x _>0+ L xl/ (c) lim [^ 1 1/3 (x - 1)V3 1 1/3 (x- 1)4/3 00 27. y = -^ - 7 X — x-1 (b) x^O- [^ (d) ^ [xi 1 xi/3 (x- I) 4 / 3 1 X l/3 (X- 1)4/3 -OO -OO 28- Y = d: x+l 2 3 4 X=- 1 y = x + 1 29- y = ^ 2x + 4 * \ 10 5 1 y 2* + 4 .*" = -2 v. -4 2 - 1 -5 -10 1 2 31. y = i±| = 1 + 4 •' x + 2 x + + 2 -2 '. \L5 y= i±3 _J_x + 2 ' **■ — . -5\ y= JT2 \ -2 30. y = -4r J x — 3 32. y 2x q 2 x+l — Z x+l 1 ' 2 A 1 y " * + 1 / ,' 1 2x y = 2 /^ — " !/ 1 x = -1 J' j ' ',-" Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 2.5 Infinite Limits and Vertical Asymptotes 95 33. y= -^ T=x+ i + -± T 34. y x^+1 _ x+1 , 2 x- 1 x- 1 _l — I — t_ "3 /yA> -2 i' x 2 1 X- 1 X- 1 12 3 4 5 35. y X— 1 X— 1 36. y x 2 -l _ 1 2x + 4 2 X- 1 2x + 4 37. y £=i=x-i 38. y x°+l 39. Here is one possibility. 40. Here is one possibility. y Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 96 Chapter 2 Limits and Continuity 41. Here is one possibility. y=m 42. Here is one possibility. y 43. Here is one possibility. 44. Here is one possibility. . , 1 5 4 3 /W ~(*-2)= 2 1 1 3 4 5 x - 3 4 e 45. Here is one possibility. 46. Here is one possibility. -> j h(x) = — , x±0 \x\ H ; V >-l 4 f kw-i-sJt ^— - — -4 -2 -2 -4 /^ 4 6 47. For every real number — B < 0, we must find a S > such that for all x, < |x — 0| < 6 =4> 4 < B. Now, - J J <-B<0o J j>B>0<^x 2 <i ■& Ixl < 4- . Choose 6 = 4- , then < Ixl < 6 => |x| < 4- x^ x^ B I I ,/b J-R I I ' ' VB -4 < -B so that lim - -4 x2 x -> x -00. 48. For every real number B > 0, we must find a 6 > such that for all x, < |x — 0| < 6 =>• A > B. Now, rr > B > O Ixl < i Choose 6 = i Then < |x - 01 < 6 => Ixl < ± => A > B so that lim A- = oo. |X| ■ ■ ±5 D ' ' ■ ■ ±5 | X| v > f) | x | For every real number B < 0, we must find a S > such that for all x, < |x — 3| < 6 (x - 3)2 < -B. Now, -2 (x - 3)2 < B < & ^^5 > B > & S-y^ < i <=> (x - 3) 2 < | ^ < \x - 3| < J I . Choose <5 = A /| , then < |x - 31 < 6 => — =% < -B < so that lim t-=|w 50. For every real number B > 0, we must find a 6 > such that for all x, < |x — (— 5)| < 6 => ,„ , 1 C „ > B. (x + 5)2 Now, (x + 5)2 > B > O (x + 5) 2 < i O |x + 5| < 4j . Choose 5 = 4g . Then < |x - (-5)| < 6 => |x + 51 < 4- => 7— 4h > B so that lim ^5 "^ (x+5)' x^ -5 (x + 5)2 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 2.5 Infinite Limits and Vertical Asymptotes 97 51. (a) We say that f(x) approaches infinity as x approaches Xo from the left, and write lim_ f(x) = oo, if X — > x for every positive number B, there exists a corresponding number 6 > such that for all x, x - S < x < x => f(x) > B. (b) We say that f(x) approaches minus infinity as x approaches Xo from the right, and write lim f(x) = — oo, x^x+ if for every positive number B (or negative number — B) there exists a corresponding number S > such that for all x, x < x < x + 6 =^ f(x) < -B. (c) We say that f(x) approaches minus infinity as x approaches Xq from the left, and write lim_ f(x) = — oo, x —* x o if for every positive number B (or negative number — B) there exists a corresponding number S > such that for all x, x - 6 < x < x =4> f(x) < -B. 52. For B > 0, - > B > <^> x < i Choose 6 = k. Then < x < <5 =4> 0< x < ^ => i > B so that lim i = oo. X D D D X ri-|- X 53. For B > 0, \ < -B < O - ± > B > 4* -x < i & - i < x. Choose <5 = i Then -6 < x < s < x => - < -B so that lim - B x x^O" x -oo. 54. For B > 0, ^ < -B ^ - ^ > B <=> -(x - 2)< A <^ x - 2 > - A «4> x > 2 - A, Choose § = A. Then 2 - <5 < x < 2 =>• -<5 < x - 2 < => -i<x-2<0 =4> ^ < -B < so that lim_ ^A_ = ~°°- 55. For B > 0, -^ > B <^> < x - 2 < ± Choose <5 = ± Then 2 < x < 2 + 6 =>• < x - 2 < 5 =4> < x - 2 < A =>- -A= > B > so that lim -A= = 00. x-z x^2+ X ~ Z 56. For B > and < x < 1, y^Uj > B 44> 1 - x 2 < A O (1- x)(l +x) < A. Now A±* < 1 since x < 1. Choose <5 < i. Then 1 - <5 < x < 1 => -<5 < x - 1 < => 1 - x < <5 < ^ =^> (1- x)(l + x) < B V 2 J v B => -r-^ > B for < x < 1 and x near 1 1 — x z 57. y = sec x + - lim i-»r 00. 58. y = sec x — \ Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 98 Chapter 2 Limits and Continuity 59. y = tan x + i 60. y = - — tan x y = tan x + 61. y W- 62. y 1 ;j: = -2 ; 2 1 ,1 ; V4-x 2 1 1 2 x=2 ; 63. y = x 2 / 3 + 4, v*~- 64. y = sin(^) y it _-^N LX 2 + 1 -i i 2.6 CONTINUITY 1 . No, discontinuous at x = 2, not defined at x = 2 2. No, discontinuous at x = 3, 1 = lim g(x) ^ g(3) =1.5 x — > 3 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 2.6 Continuity 99 3. Continuous on [— 1, 3] 4. No, discontinuous at x = 1, 1.5 = lim k(x) ^ lim k(x) = x -> 1 X -> 1+ 5. (a) Yes (c) Yes (b) Yes, lim f(x) = X — > — 1 + (d) Yes 6. (a) Yes, f(l) = 1 (c) No (b) Yes, lim f(x) = 2 X — > 1 (d) No 7. (a) No (b) No 8. [-1,0)U(0,1)U(1,2)U(2,3) 9. f(2) = 0, since lim f(x) = -2(2) + 4 = 0= lim f(x) x -> 2 x -> 2+ 10. f( 1 ) should be changed to 2 = lim f(x) X — > 1 11. Nonremovable discontinuity at x = 1 because lim f(x) fails to exist ( lim f(x) = 1 and lim f(x) = 0). x -> 1 x^l x^l + Removable discontinuity at x = by assigning the number lim f(x) = to be the value of f(0) rather than x^O f(0) = 1. 12. Nonremovable discontinuity at x = 1 because lim f(x) fails to exist ( lim f(x) = 2 and lim f(x) = 1). X -> 1 X -» 1 X -> 1+ Removable discontinuity at x = 2 by assigning the number lim f(x) = 1 to be the value of f(2) rather than x — > 2 f(2) = 2. 13. Discontinuous only when x — 2 = => x = 2 14. Discontinuous only when (x + 2) 2 = => x = — : 15. Discontinuous only when x 2 — 4x + 3 = =4> (x — 3)(x — 1) = =4> x = 3orx=l 16. Discontinuous only when x 2 — 3x — 10 = => (x — 5)(x + 2) = =>■ x = 5 or x = — 2 17. Continuous everywhere. ( |x — 1| + sin x defined for all x; limits exist and are equal to function values.) 18. Continuous everywhere. ( |x| + 1 ^ for all x; limits exist and are equal to function values.) 19. Discontinuous only at x = 20. Discontinuous at odd integer multiples of |, i.e., x = (2n — 1) |, n an integer, but continuous at all other x. 21. Discontinuous when 2x is an integer multiple of -it, i.e., 2x = rnr, n an integer => x= y,nan integer, but continuous at all other x. 22. Discontinuous when y is an odd integer multiple of |, i.e., y = (2n — l) |, n an integer => x = 2n — 1, n an integer (i.e., x is an odd integer). Continuous everywhere else. Cop # (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 100 Chapter 2 Limits and Continuity 23. Discontinuous at odd integer multiples of |, i.e., x = (2n — 1) |, n an integer, but continuous at all other x. 24. Continuous everywhere since x 4 + 1 > 1 and — 1 < sin x < 1 =>• < sin 2 x < 1 => 1 + sin 2 x > 1 ; limits exist and are equal to the function values. 25. Discontinuous when 2x + 3<0orx<— | =>• continuous on the interval [— |, oo) . 26. Discontinuous when 3x— l<0orx<j =4> continuous on the interval [i, oo) . 27. Continuous everywhere: (2x — l) 1 / 3 is defined for all x; limits exist and are equal to function values. 28. Continuous everywhere: (2 — x) 1 / 5 is defined for all x; limits exist and are equal to function values. 29. lim sin (x — sin x) = sin (it — sin it) = sin (it — 0) = sin it = 0, and function continuous at x = it. X — > w 30. lim sin (| cos (tan t)) = sin (| cos (tan (0))) = sin (| cos (0)) = sin (|) = 1, and function continuous at t = 0. 31. lim sec (y sec 2 y — tan 2 y — 1) = lim sec (y sec 2 y — sec 2 y) = lim sec ((y — l)sec 2 y) = sec((l — l)sec 2 1) y -> 1 y -> 1 " y -» 1 = sec 0=1,, and function continuous at y = 1. 32. lim tan [| cos (sin x 1 / 3 )] = tan [| cos (sin(0))] = tan (? cos (0)) = tan (|) = 1, and function continuous at x = 0. 33. lim cos , - t— »o |_yi9-3sec2t i — cos -t— = cos ? = o > and function continuous at t = 0. V19-3SBC0J \/16 4 2 34. lirn. -i/csc 2 x + 5v^3 tan x = Jcsc 2 (|) + 5\/3 tan (|) = w4 + 5^/3 (4- J = y^9 = 3, and function continuous at <;• 35. g(x) = fff = (x + x 3Kx 3 ^ 3) = x + 3, x + 3 =► g(3) = x lim 3 (x + 3) = 6 36. h(t) = t2 + 3t 2 1 ° = (t+ t 5 l (t 2 ~ 2) = t + 5, t ^ 2 =>■ h(2) = lim (t + 5) = 7 37. f(s) -1 _ (s 2 + s+ l)(s- 1) _ sj+s + 1 s 2 -! (s + l)(s-l) *£±Wl ^f(l) = s lim i (^±l) 38. g(x) = -#^ = f + Tln = ^tt • x ¥= 4 =5- g(4) = lim (^±4) = | &v ' x 2 — 3x — 4 (x — 4)(x+l) x+1 ' * b\ i „ _^ a \x+l/ 5 39. As defined, lim_ f(x) = (3) 2 — 1 = 8 and lim (2a)(3) = 6a. For f(x) to be continuous we must have x — > 3 x -> 3+ 6a = 8 =^ a = \. 40. As defined, lim g(x) = —2 and lim g(x) = b(— 2) 2 = 4b. For g(x) to be continuous we must have 4b = -2 => b = -i Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 2.6 Continuity 101 41. The function can be extended: f(0) ~ 2.3. 42. The function cannot be extended to be continuous at x = 0. If f(0) « 2.3, it will be continuous from the right. Or if f (0) w —2.3, it will be continuous from the left. -0.1 -0.05 fix)' 10 W - 1 0.05 0.1 43. The function cannot be extended to be continuous at x = 0. If f(0) = 1 , it will be continuous from the right. Or if f(0) = — 1 , it will be continuous from the left. 44. The function can be extended: f(0) ~ 7.39. l(r 0.5 -0.1 -0.05 -0.5 i 0.05 0.1 /(*) = sin* -0.01 -0.005 fix) = (1 + 2*)"* 0.005 0.01 45. f(x) is continuous on [0, 1] and f(0) < 0, f(l) > =^ by the Intermediate Value Theorem f(x) takes on every value between f(0) and f(l) => the equation f(x) = has at least one solution between x = and x = 1 . 46. cos x (cos x) — x 0. Ifx=-§, cos (-§)-(-§) > 0. If x COS I for some x between — | and | according to the Intermediate Value Theorem. — | < 0. Thus cos x — x = 47. Let f(x) = x 3 - 15x + 1 which is continuous on [-4, 4]. Then f(-4) = -3, f(- 1) = 15, f(l) = - 13, and f(4) = 5. By the Intermediate Value Theorem, f(x) = for some x in each of the intervals —4 < x < — 1, — 1 < x < 1, and 1 < x < 4. That is, x 3 — 15x +1=0 has three solutions in [—4, 4]. Since a polynomial of degree 3 can have at most 3 solutions, these are the only solutions. Without loss of generality, assume that a < b. Then F(x) = (x — a) 2 (x - x, so it is continuous on the interval [a, b]. Moreover F(a) = a and F(b) b) 2 + x is continuous for all values of = b. By the Intermediate Value Theorem, since a < ^4^ < b, there is a number c between a and b such that F(x) Copyright (c) 2006 Pearson Education 102 Chapter 2 Limits and Continuity 49. Answers may vary. Note that f is continuous for every value of x. (a) f(0) = 10, f(l) = l 3 - 8(1) + 10 = 3. Since 3 < 7r < 10, by the Intermediate Value Theorem, there exists a c so that < c < 1 and f(c) = n. (b) f(0) = 10, f(-4) = (-4) 3 - 8(-4) + 10 = -22. Since -22 < -x/3 < 10, by the Intermediate Value Theorem, there exists a c so that —4 < c < and f(c) = — \J 3. (c) f(0) = 10, f(1000) = (1000) 3 - 8(1000) + 10 = 999,992,010. Since 10 < 5,000,000 < 999,992,010, by the Intermediate Value Theorem, there exists a c so that < c < 1000 and f(c) = 5,000,000. 50. All five statements ask for the same information because of the intermediate value property of continuous functions. (a) A root of f(x) = x 3 — 3x — 1 is a point c where f(c) = 0. (b) The points where y = x 3 crosses y = 3x + 1 have the same y-coordinate, or y = x 3 = 3x + 1 => f(x) = x 3 - 3x - 1 = 0. (c) x 3 — 3x = 1 =4> x 3 — 3x — 1 = 0. The solutions to the equation are the roots of f(x) = x 3 — 3x — 1 . (d) The points where y = x 3 — 3x crosses y = 1 have common y-coordinates, or y = x 3 — 3x = 1 => f(x) = x 3 - 3x - 1 = 0. (e) The solutions of x 3 — 3x — 1 = are those points where f(x) = x 3 — 3x — 1 has value 0. 51. Answers may vary. For example, f(x) = — _~ ' is discontinuous at x = 2 because it is not defined there. However, the discontinuity can be removed because f has a limit (namely 1) as x — > 2. 52. Answers may vary. For example, g(x) = —j-t has a discontinuity at x = —1 because lim g(x) does not exist. ( lim g(x) = — oo and lim g(x) = +oo. J \x->-l x ^_l+ / 53. (a) Suppose Xo is rational =$■ f(xo) = 1. Choose e = \. For any 6 > there is an irrational number x (actually infinitely many) in the interval (xo — 6, Xo + S) =>■ f(x) = 0. Then < [x — Xoj < S but |f(x) — f(xo)| = 1 > I = e, so lim f(x) fails to exist =>- f is discontinuous at Xn rational. i X — > X On the other hand, Xo irrational =^ f(xo) = and there is a rational number x in (xo — 6, Xq + S) =^ f(x) = 1. Again lim f(x) fails to exist =>- f is discontinuous at Xn irrational. That is, f is discontinuous at X — > x every point, (b) f is neither right-continuous nor left-continuous at any point x because in every interval (x — 6, x ) or (xo, Xo + S) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and x — > x lim f(x) exist by the same arguments used in part (a). 54. Yes. Both f(x) = x and g(x) = x — | are continuous on [0, 1]. However -£i is undefined at x = | since g(i)=0 => -^ is discontinuous at x = 1. 55. No. For instance, if f(x) = 0, g(x) = |~x] , then h(x) = ( |~x] ) = is continuous at x = and g(x) is not. 56. Let f(x) = ^y and g(x) = x + 1. Both functions are continuous at x = 0. The composition f o g = f(g(x)) = , j._ l = i is discontinuous at x = 0, since it is not defined there. Theorem 10 requires that f(x) be continuous at g(0), which is not the case here since g(0) = 1 and f is undefined at 1. 57. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to equal zero at some point between a and b since f is continuous on [a. b]. Copfiigl (c) 1 Pearson Etation, Inc., publishing as Pearson Addison-Wesle Section 2.7 Tangents and Derivatives 103 58. Let f(x) be the new position of point x and let d(x) = f(x) — x. The displacement function d is negative if x is the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d(x) = for some point in between. That is, f(x) = x for some point x, which is then in its original position. 59. If f(0) = or f(l) = 1, we are done (i.e., c = or c = 1 in those cases). Then let f(0) = a > and f(l) = b < 1 because < f(x) < 1. Define g(x) = f(x) — x =► g is continuous on [0, 1]. Moreover, g(0) = f(0) — = a > and g(l) = f(l) — 1 = b — 1<0 =4> by the Intermediate Value Theorem there is a number c in (0, 1) such that g (c) = => f(c) - c = or f(c) = c. 60. Lete |f«0l > 0. Since f is continuous at x = c there is a 6 > such that |x — c| < 6 =4> |f(x) — f(c)| < e => f(c) - e < f(x) < f(c) + e. If f(c) > 0, then e = \ f(c) => \ f(c) < f(x) < § f(c) =>■ f(x) > on the interval (c - 6, c + 6). If f(c) < 0, then e = - \ f(c) =^ § f(c) < f(x) < \ f(c) => f(x) < on the interval (c - 6, c + S). f(c>- C-<5 C+6 f(c)+e 61. By Exercises 52 in Section 2.3, we have lim f(x) = L <^> lim f(c + h) = L. x — > c h — > Thus, f(x) is continuous at x = c •<=>■ lim f(x) = f(c) <=> lim f(c + h) = f(c). x — > c h _> o 62. By Exercise 61, it suffices to show that lim sin(c + h) = sin c and lim cos(c + h) = cos c. J h->0 v ' h-»0 v ' Now lim sin(c + h) = lim [(sin c)(cos h) + (cos c)(sin h)l =(sinc)( lim cos h ) + (cos c)( lim sin h) By Example 6 Section 2.2, lim cos h = 1 and lim sin h = 0. So lim sin(c + h) = sin c and thus f(x) = sin x is continuous at x = c. Similarly, lim cos(c + h) = lim [(cos c)(cos h) — (sin c)(sin h)l = (cos c) I lim cos h ) — (sin c) ( lim sin h ) = cos c. Thus, g(x) = cos x is continuous at x = c. 63. x w 1.8794, -1.5321, -0.3473 64. x sa 1.4516, -0.8547, 0.4030 65. x w 1.7549 67. x« 3.5156 69. x« 0.7391 66. xw 1.5596 68. x w -3.9058, 3.8392, 0.0667 70. xw -1.8955,0, 1.8955 2.7 TANGENTS AND DERIVATIVES 1. P i: mi = 1,P 2 : m 2 = 5 2. Pi: mi = -2, P 2 : m 2 = Cop # (c) 1 Pen %$A\ k, publishing as Pearson Addison-Wesle 104 Chapter 2 Limits and Continuity 3. Pi: mi = |,P 2 : m 2 4. Pi: mi = 3, P 2 : m 2 = -3 5. m = lim [4-(-l + h) 2 ]-(4-(-l) 2 ) lim h->0 (l-2h + h 2 ) + l lim h->0 h(2 - h) at (-1,3): y = 3 + 2(x-(-l)) => y = 2x + 5, tangent line y h y = 2x + 5 / C (-1,3)/ 3 2 1 V = 4-* 2 -3 / /-2 -1 I 2 \ 6. m = lim " 1+h - 1)2+1 i-' (1 - 1)2+ " = lim h-' h->0 h-»0 lim h = 0; at (1, 1): y = 1 + 0(x - 1) =4> y h — > tangent line 5 \ * \ 3 y.(x-1) 2 + 1 / (i.i) y-1 -1 -1 1 2 7. m = lim h->0 2y/l+h-2yi h 4(l+h)-4 lim h^O 2h(%/l+h+l) lim h-»0 lim 2yi+h-2 2y/l+h + 2 2vT+h + 2 = 1; h-»0 Vl+h+l at (1,2): y = 2 + l(x - 1) =>• y = x + 1, tangent line , y = x + \/ 4 3 A/y = l-{x 2 1 - y<i,2) 12 3 4 J 1 m= lim ±±±^ h->0 i-(-i+hy lim — — r^^- h _> o h(-!+ h ) 2 -(-2h + h 2 ) lim h ^0 h (-!+ h ) 2 lim 2-h h ^o (- 1 + h > 2 ' at (-1, 1): y = 1 + 2(x - (-1)) => y = 2x + 3, tangent line Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 2.7 Tangents and Derivatives 105 9. m = lim h->0 (-2 + h) 3 -(-2) 3 lim h->0 -8+12h-6h 2 + h 3 + 8 = lim (12-6h + h 2 ) = 12; IwO v ' at (-2, -8): y = -8 + 12(x - (-2)) tangent line y = 12x+ 16, y=12*+16 10. m= lim '- 2+h]3 h '- 2 ' 3 h->0 h -8-(-2 + h) 3 h ™ -8h(-2 + h) 3 j. -(12h-6h 2 + h 3 ) _ ,. 12-6h + tf hl^O -8h(-2 + h) 3 _ h lml 8(-2 + h) 3 12 _3_. 16' at (-2, -I): y =-I_^(x-(-2)) =>■ y = — fg x — |, tangent line (-2,-1/8) 16 2 11. m = lim [P + ^ + H-s = lim (5 + 4h + hV5 = Um h(4 + h) = 4 h -> h h -» h h -> h at (2, 5): y - 5 = 4(x - 2), tangent line 12. m = lim 1(1 + h) - 2(1 h +h)2] - ( - 1) = lim il±!l=2=*=2*l±l = lim Wh) = _ 3 h->0 h h->0 h IwO h at (1, —1): y + 1 = — 3(x — 1), tangent line 13. m= lim 2±t 3+h o 13 + h) — 2 h->0 lim (3 + h)-3(h+l) lim -2h h — h(h+l) h — h(h+l) -2; at (3, 3): y — 3 = — 2(x — 3), tangent line 14. m = lim ^^ = lim ^±^ = lim ^±g£^ = lim ^i+h) _ ^ h->0 h ^* h(2 + h)2 h^o h(2 + h)2 _ h "o h (2 + h)2 - 4 -2; at (2, 2): y - 2 = -2(x - 2) 15. m= lim V±t^= lim (« + ^ + f + "3)-g = ^ h(i2 +6 h + tf) = 12 h->0 h->0 h->0 at (2, 8): y - 8 = 12(t - 2), tangent line 16. m = lim ' (1+h)3 + 3 h (1+h) '- 4 = lim (' + 3h + 3h 2 + h 3 + 3 + 3h)-4 = ^ h(6 + 3h + h 2 ) = h->0 h h->0 h h->0 h at (1, 4): y - 4 = 6(t - 1), tangent line lim l n i- V4 + h-2 ,. V4 + h-2 V4 + 11 + 2 ,■ (4 + h)-4 17. m = lim - — s = lim - — r • v , — = lim / , ; — ^ — "■" / , — \- — —r. — h->0 h h->0 h \/4 + h + 2 h^O h(v/4 + h + 2j h -> h(y4 + h + 2j v / 4 + 2 |; at (4, 2): y — 2 = \ (x - 4), tangent line lim 18. m = um v^+h)TT-3 = j. m ^+h-3 . ^+3 = lim (9 + h)-9 _ mii ____ h->0 h h-»0 h V9 + h + 3 h-»0 h(y9 + h + 3J h^O h( v /9 + h + 3j = 7^ = |; at (8,3): y - 3 = | (x - 8), tangent line 19. Atx=-l,y = 5 =4> m = lim 5(-l+h) 2 -5 h lim 5( '- 2h h +h2) - 5 = lim 5h <4±W = -10, slope h^O h h-»0 h V Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 106 Chapter 2 Limits and Continuity 20. Atx = 2,y = -3 => m= lim Ii^±^Mz5 = i im (i-4-4h-h 2 ) + 3 = Um zM 4 + h) = _ 4 j h->0 h h->0 h IwO h 2 - (2 + h) 21. At x = 3, y = i =4> m = lim ,3+h> -' 2 = lim tT^T^' = lim ,, , , , - ' J 2 h^O h h^O 2h<2 + h) h -> 2h(2 + h) 4 I, slope 22. At x = 0, y = -1 => m = lim ' :+l . ( ' } = lim (h ~'^ + 1} = lim z^-r, = 2, slope J h-0 h h->0 h(h+1 » h->0 h(h+1) F 23. At a horizontal tangent the slope m = => = m= lim K* + h > 2 + 4 <* + h >- i]-(x 2 + 4x- l) lim (* 2 + 2 *h + h 2 + 4x + 4h-l)-(x 2 + 4x-l) =; Hm (2xh + h !± 4h) = ^ (2x + h + 4) = 2x + 4; h-»0 h->0 h->0 2x + 4 = => x= —2. Then f(— 2) = 4 — 8— 1 = — 5 =>• (—2, —5) is the point on the graph where there is a horizontal tangent. 24. = m = lim h->0 [(x + h) 3 - 3(x + h)] - (x 3 - 3x) __ ,. (x 3 + 3x 2 h + 3xh 2 + h 3 - 3x - 3h) - (x 3 - 3x) h->0 = lim 3x 2 h + 3xh 2 + h 3 -3h = Hm / 3x 2 + 3 x h + h 2 - 3) = 3x 2 - 3; 3x 2 - 3 = =>■ x = -1 or x = 1. Then h->0 h IwO v ' f(— 1) = 2 and f(l) = — 2 =>• (—1,2) and (1, —2) are the points on the graph where a horizontal tangent exists. lim (x-l)-(x + h-l) lim -h 1 l i 95 i — t-,-, — Htti <x+1 "~' i=- - ,,,,, - J ' hn o h ^"f, h(x-l)(x + h-l) h "^" h(x-l)(x + h-l) (x-1) 2 v2 _ 1 _^ v 2 => (x - l) 2 = 1 => x 2 - 2x = => x(x - 2) = =>• x = or x = 2. If x = 0, then y = -1 andm = -1 =>■ y=-l-(x-0) = -(x + 1). Ifx = 2, theny = 1 andm = -1 =4> y = 1 - (x - 2) = -(x - 3). r,r \ i- \Jx + h- \A l- \/x + h - ,/x v^ x + h + \/x ,. (x + h)-x 26. 4 = m = lim c — — = hm c — — • ; , — ^ = lim h->0 h^O /x + h+^/x h^O hK/x + h + nA) lim h^O h^Vx + h+yj) 2, A y = 2+i(x-4)=| + l ^7= . Thus, I = -^-/= =>■ a/x = 2 =>• x = 4 =>• y = 2. The tangent line is 2^/1 2? lim f(2 + h)-f(2) = j im (100-4.9(2 + h) 2 )-(100-4.9(2) 2 ) = ^ -4.9 (4 + 4h + h 2 ) +4.9(4) ' h^O h h->0 h h->0 h = lim (—19.6 — 4.9h) = — 19.6. The minus sign indicates the object is falling downward at a speed of h — > 19.6 m/sec. 28. lim nio + h)-f(io) = lim 3 ( io + h) 2 -3(io) 2 = Hm 3(201^) =60ft/sec _ h->0 h h->0 h h^O h 29. lim '' ^^= lim " (3 + h) ! ~ " (3)2 = , lim, *P+6M*»-9] = lim ^ (6 + h) = 6 ^ h->0 30. lim h->0 h f(3 + h)- •f(3) h f(2 + h) - f(2) h^O lim h->0 h->0 h->0 4 F(2 + h) 3 -fe(2) 3 _ f [12h + 6h 2 + h 3 ] _ 47r f(0 + h)-f(0) 31. Slope at origin = lim v b h->0 h the origin with slope 0. lim h-»0 lim h 2 sin(l) lim % fl2 + 6h + h 2 l = 16tt h^ 3 l J lim hsin(^) =0 =>• yes, f(x) does have a tangent at 32. lim g( + I — — = lim T = lim sin ^ Since lim sin r does not exist, f(x) has no tangent at h->0 the origin. h->0 h->0 h->0 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 33. lim h-»0~ f(o + h)-f(o )= Hm h twO" lim h->0 f(0 + h)-f(0) =oo ^ 34. lim h-»0~ u(o + h)-u(0 )= Hm h h-»0 =t=fi = oo,and lim f(0 + h >- f(0) = lim I=° = h h->0+ h h ^0+ h oo =>• yes, the graph of f has a vertical tangent at the origin. Section 2.7 Tangents and Derivatives lim i -^- y = oo. Therefore, 107 V = oo,and lim u(0 + h > ~ u(0) h h->0+ h lim h-»0+ l- l h =>- no, the graph of f does not have a vertical tangent at (0, 1) because the limit does not exist. 35. (a) The graph appears to have a cusp at x = 0. (b) lim h -> 0- f(0 + h)-f(0) lim h-»(r h 2 /=-0 lim h 3/5 -oo and lim h->0+ h 3 5 limit does not exist the graph of y = x 2 / 5 does not have a vertical tangent at x = 0. 36. (a) The graph appears to have a cusp at x = 0. (b ) Hm ffl + h)-f(0) = lim h -> 0" h h -> 0" h 4 / 5 - lim tjw h^0" hl/o — oo and lim lw0+ hV5 oo =>- limit does not exist y = x 4 / 5 does not have a vertical tangent at x = 0. 37. (a) The graph appears to have a vertical tangent at x = 0. (b) lim go + h)-f(0) = Hm hV^o = Hm h->0 h h->0 h h->0 h 4 / 5 (0,0) y = x 1 / 5 has a vertical tangent at x = 0. 38. (a) The graph appears to have a vertical tangent at x = 0. (0,0) y =x 3/5 (b) lim f (° + h > f W = ij m h3/o ° = Hm -^ = 00^ the graph of y = x 3/5 has a vertical tangent w h-0 h h->0 h h-u bV ° & F J & at x = 0. Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 108 Chapter 2 Limits and Continuity 39. (a) The graph appears to have a cusp at x = 0. y = 4x 2/5 -2x (b ) lim f(0 + h >- fW = lim h -> 0" h h -> 0" 4h 2 / 5 - 2h lim h->0" h 3 .:, -co and lim h-»0+ h 3 .-. 2 = oo =>• limit does not exist =4> the graph of y = 4x 2 / 5 — 2x does not have a vertical tangent at x = 0. 40. (a) The graph appears to have a cusp at x = 0. 5/3 - 2/3 y ■ x - 5x (b) lim h->0 f(0 + h) - f(0) lim h-»0 h 5/3_ 5h 2/3 lim h 2 / 3 h->0 7-yj = — lim j-^3 does not exist =4> the graph of y = x 5 / 3 — 5x 2 / 3 does not have a vertical tangent at x = 0. 41. (a) The graph appears to have a vertical tangent at x = 1 and a cusp at x = 0. y = x 2 / 3 -(x-l) 1 / 3 (b) x = 1: lim d+h^-d+h- i)i/3-i h-»0 h lim h->0 ( 1 + h) 2 /' 3 - h 1 / 3 - 1 y = x 2 / 3 — (x — l) 1 ' 3 has a vertical tangent at x = 1; x = 0: lim h-»0 gO + h) - f(0) lim h->0 h2/3-( h - 1)V3 -(-1)1/3 lim h->0 (h-1) 1 ' 3 , 1 h 13 does not exist =^ y = x 2//3 — (x — l) 1 / 3 does not have a vertical tangent at x = 0. Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 2.7 Tangents and Derivatives 109 42. (a) The graph appears to have vertical tangents at x = and x= 1. (b ) x = 0: lim f(0 + h '- f(0) h-0 h lim hl/3 + <h - 1 > 1/3 -^ 1 > 1/3 =oo h->0 h vertical tangent at x = 0; x = 1: lim h->0 f(l+h)-f(l) lim h->0 vertical tangent at x = 1. (1 + h) 1 / 3 + (l +h— l) 1 / 3 — 1 '0.5 1.5 y-x 1/3 *(x-l)l'3 y = x 1 / 3 + (x - l) 1 / 3 has ; oo => y = x 1 / 3 + (x - l) 1 / 3 has a 43. (a) The graph appears to have a vertical tangent at x = 0. (b) lim f(0 + h h ) - f(Q) W h->0+ h lim h-v(T f(0 + h)-f(0) lim lim h->0~ y/h-0 h h lim -T- = oo; h->0 Vh lim =%$■ lim h->0~ =>- y has a vertical tangent at x = 0. 44. (a) The graph appears to have a cusp at x = 4. (b ) lim f(4 + h >- f(4) lim lim h->CT => y f(4 + h)-f(4) lim h->(T VK- -(4 + h)|- -0 h VK- -(4 + h)| OO lim h^0+ lim -4- = oo; h^o+ v h lim h h -» o- h V 4 — x does not have a vertical tangent at x = 4. lim h-»tr 45-48. Example CAS commands: Maple : f := x -> x A 3 + 2*x;x0 := 0; plot( f(x), x=x0- 1/2..X0+3, color=black, # part (a) title="Section 2.7, #45(a)" ); q := unapply( (f(x0+h)-f(x0))/h, h ); # part (b) L := limit( q(h), h=0 ); # part (c) secjines := seq( f(xO)+q(h)*(x-xO), h=1..3 ); # part (d) tanjine := f(xO) + L*(x-xO); plot( [f(x),tan_line, secjines], x=x0-l/2..x0+3, color=black, Cfiojt (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle 110 Chapter 2 Limits and Continuity linestyle=[l,2,5,6,7], title=" Section 2.7, #45 (d)", legend=["y=f(x)", "Tangent line at x=0", "Secant line (h=l)", "Secant line (h=2)", "Secant line (h=3)"] ); Mathematica : (function and value for xO may change) Clear[f, m, x, h] x0 = p; f[x_]: = Cos[x] + 4Sin[2x] Plot[f[x],{x,x0- l,x0 + 3}] dq[h_]: = (f[x0+h] - f[x0])/h m = Limit[dq[h],h -> 0] ytan: = f[x0] + m(x - xO) yi y2 y3 f[x0] + dq[l](x - xO) f[x0] + dq[2](x - xO) f[x0] + dq[3](x - xO) Plot[{f[x],ytan,yl,y2,y3}, {x,xO - l,xO + 3}] CHAPTER 2 PRACTICE EXERCISES 1. Atx f(x) lim f(x)=l=f(-l) 1: lim x->-l lim f(x) x->-l f is continuous at x 1. At x = 0: lim f(x) = lim f(x) = x -» 0" x -» 0+ Butf(0) = 1/ lim f(x) x — > lim f(x) = 0. x — » => f is discontinuous at x = 0. If we define f(0) = 0, then the discontinuity at x = is removable. Atx = 1: lim f(x) X — > 1 1 and lim f(x) = 1 x^ 1+ lim f(x) does not exist x — > 1 f is discontinuous at x = 1 . 2. At x = -1: lim _ f(x) = and lim f(x) x — > -1 X -» -1+ =>• lim f(x) does not exist X — > —1 =>• f is discontinuous at x = — 1 . Atx = 0: lim_f(x) x — > -oo and lim f(x) x^0+ =>- lim f(x) does not exist x — > =4> f is discontinuous at x = 0. Atx= 1: lim f(x) = lim f(x) -- x -> I" X -> 1+ Butf(l) = 0^ lim f(x) X — » 1 lim f(x) = 1. X — > 1 =4> f is discontinuous at x = 1 . If we define f(l) = 1, then the discontinuity at x = 1 is removable. A l<- y=/« 1 -1 -1 1 1 -*-x /(*) = l/x, < \x\ < 1 0, * = 1 1. x> I i 1 -1 V l 3. (a) lim (3f(t)) = 3 lim f(t) = 3(-7) = -21 (b) lim (f(t)r t -^ to ( t lim f(t)) 2 = (-7) 2 = 49 Copyright (c) 2006 Pearson Education Chapter 2 Practice Exercises 111 (c) lim (f(t) • g(t)) = lim f(t) • lim g(t) = (-7)(0) = t — > tfl t — > to t — > to (d) lim t-»to f(0 g(t)-7 lim ft) t — »tQ lim (g(t) - t— *to lim f(t) lim o(t) — Hm7 t-rto t->to -7 0-7 (e) lim cos (g(t)) = cos ( lim g(t) ) = cosO = 1 t — » to ' \t — > t / 'lim f(t)| = |-7|=7 (f) lim |f(t)| = t — > to i i —i- K) l (g) lim (f(t) + g(t)) = lim f(t) t — * to t — > t lim g(t) t — » to (h) lim t->to (A) lim ft) t— »Q 4. (a) lim -g(x) x — > - lim g(x) x — » (b) lim (g(x) • f(x)) = lim g(x) - lim f(x) = (y/l) (§} x^O x^O x^O V / (c) lim (f(x) + g(x)) = lim f(x) + lim g(x) x — > x^O x^O (d) lim ± - —^- - i - t V ' x^O f W lim f(x) x^o (e) lim (x + f(x)) = lim x + lim f(x) = - (f) Hm n x — > f(x)-cos X x-1 lim f(x)- lim cos x x^O x^O lim x — lim 1 x-tO x^O 0-1 5. Since lim x = we must have that lim (4 — g(x)) = 0. Otherwise, if lim (4 — g(x)) is a finite positive x^O x^O x^O number, we would have lim x->(r 4-g(x) -oo and lim x^0 + 4-g(x) oo so the limit could not equal 1 as x — ■> 0. Similar reasoning holds if lim (4 — g(x)) is a finite negative number. We conclude that lim g(x) = 4. x — > x — > 6. 2 = lim x lim g(x) = lim x - lim lim g(x) = -4 lim lim g(x) = -4 lim g(x) x — > — 4 L x — > J x^-4 x ^ -4 Lx ^ J x-»-4 1x^1) J x — > (since lim g(x) is a constant) =>■ lim g(x) = ^ = — \ . x^O x^O 7. (a) lim f(x) = lim x 1 / 3 = c 1 / 3 = f(c) for every real number c => f is continuous on ( — oo, oo). (b) lim g(x) = lim x 3 ' 4 = c 3 ' 4 = g(c) for every nonnegative real number c => g is continuous on [0, oo). (c) lim h(x) = lim x -2 / 3 = i = h(c) for every nonzero real number c =>• h is continuous on (— oo, 0) and (— oo, oo). (d) lim k(x) = lim x~ 1//(3 = i = k(c) for every positive real number c =^> k is continuous on (0, oo) 8- ( a ) U (( n — I) 71 "; ( n + I) 71 ") ' wnere I = me set of all integers. nei (b) U ( n7r J ( n + l) 71 ")' where I = the set of all integers. nei (c) ( — oo, 7r) U (w, oo) (d) (-oo, 0) U (0, oo) (x - 2)(x - 2) lim x_>0- x ( x + 7 ' x 2 - 4x + 4 oo and lim x^O 4 (b) x hm 2 x 3 + 5x 2 _ 14x — ^^2 x(x + 7)(x-2) lim x(x + 7) (x-2)(x-2) lim -00 x-2 9. (a) ^lirr^ x 3 + 5x 2_ 14x - x lim o x(x + 7)(x _ 2 ) _ x "_^" x(x + 7) x^2; the limit does not exist because lim f ,~. ,x^2, and lim -^ x(x + 7) ' ' ' x 9 x(x x-2 + 7) 2(9) 10. (a) lim * +1 3 v ' x — > + + Now lim lim x(x+l) lim x+l -q x 3 (x 2 + 2x+l) x^To * a (*+l)(*+l) lim x 2 (x+l) ,x/0 and x ^ — 1 . 0- x2 ("+l) ooand x !IJV P ^+ T > lim x 5 + 2x 4 - x 3 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 112 Chapter 2 Limits and Continuity x(x+l) lim 1 (b) x lim i x 5 + X 2x 4 X +x 3 - x ^ m 1 X 3(x2 + 2x+l) - x ^ > "i 1 X 2 (x+1) 1 x / and x^-1. The limit does not exist because lim r . — — oo and lim .-,. 1 , , , = oo. j- x 2 (x+l) x _^ _ 1+ x 2 (x+l) 11. lim V^ = lim l-x/x" lim x^l '-* x^l (l-V^)(l + v^) x^l 1 + \A 2 12 ' x lml a xV^I = x lyjl a ( x 2 + a 2 )(x 2_ a 2) = x lim a ^2^^2 = 2? 13. Hm (x + h) 2 -x 2 = lim ( x2 + 2hx + h 2 )-x 2 = Hm (2x + h) = 2x h->0 h h->0 h h->0 14. lim (x + by - x' lim x^O (x 2 + 2hx + h 2 ) - x 2 lim (2x + h) = h x — > _j l 15 Hm 2±* i _ ,„„ _ ,,,,, x^O x x^O 2x(2 + x) x ^g 4 + 2x Um ^±4 = lim - 1 16. Hm U±*Lll = Hm (x3 + 6x 2 + 12x + 8)-8 = ^ , 2 fe j 2) = 12 x^O x x^O x x -> v 17. lim [4 g(x)] 1 / 3 = 2 =>• [lim 4 g(x)l = 2 =>■ lim 4 g(x) = 8, since 2 3 = 8. Then lim g(x) = 2. x — > 0+ Lx — » 0+ J x — > 0+ x — > + 18. lim xT ^- = 2 => lim (x + g(x)) = \ => y/S + lim g(x) = \ =► lim g(x) = \ - ^5 Y 1 a / .^i a Y i -> / P\ Y V ■. / S Y 1 . / *! x^ x/5 x^ V^ 3x 2 + 1 19. lim ^Y- = oo =>• lim g(x) = since lim (3x 2 + 1) = 4 x^ 1 x^ 1 x -> 1 sOO 20. lim ^™- = =4> lim g(x) = oo since lim (5 - x 2 ) = 1 21. lim 2x + 3 lim 2 + i _ 2 + _ 2 x ^ oo 5x + 7 x^oos + i 5 + 5 22. lim 2x 2 + 3 lim ^±1- 2 + o_2 oo 5x 2 + 7 x — > -oo 5+4 5 + ° 5 23. lim ^^ x — » -oo 3x J x ijm oc (i-^ + 3| 3 ) = o-o + o = o 24. lim ^-4 lim o x^ocx 2 -7x + l x^ool- 2 + 4, 1-0 + 25. lim x"-7x 00 x+1 lim *=X — > —001 + 7 -oc 26. lim . 'o 4 ; ' . = lim x^oo 12x3 +128 x _^ 1 ii ool2 + l 2 8 oc 27. lim ^-r^ < lim r^ = since int x — > oo as x — » oo => lim ^r 11 ^ = 0. X^CO W — X — > oo L X J X — » 00 L X J 28. lim ^^ < lim ^ = => lim ^^ = 0. 29. lim x + sin x + 2 */x lim x ^ oo x + sinx x ^ oo 1 + "7^ _ 1+0+0 _ i * i + o x 30. lim x2/3+x " 1 lim x — > oo x z / J + cos^x x — > oo \ i 1 + _ i 1+0 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Chapter 2 Practice Exercises 113 31. Atx 1: lim f(x) X — » — 1 lim x(x 2 -l) 1x2-1 lim -1- x(x 2 -l) X 2-l lim _ x = — 1, and lim f(x) lim x(x 2 -l) lim x(x 2 -l) -(X2-1) 1+ I* 2 - 'I X--1+ = lim (-x) = -(-1) = 1. Since X — > — 1 lim _ f(x) ^ lim f(x) x —> -1 X —> -1+ =>• lim f(x) does not exist, the function f cannot be X — > — 1 extended to a continuous function at x = — 1 . /C*)-;r(* , -l)/|* , -l| At x = 1 : lim f(x) = lim X — > 1 X — > 1 l x 'I lim (-x) X — > 1 - 1 , and lim f(x) = lim ^4 — rp lim x(x 2 -l) x 2 -l lim x = 1 . Again lim f(x) does not exist so f ; -> 1+ x-»l cannot be extended to a continuous function at x = 1 either. 32. The discontinuity at x = of f(x) = sin (-) is nonremovable because lim sin - does not exist. \X/ X — > x 33. Yes, f does have a continuous extension to a = 1: define f(l)= lim ^ x^ 1 x- 34. Yes, g does have a continuous extension to a lim 5 cosg 40 - 2tt 5 4' 35. From the graph we see that lim h(t) ^= lim h(t) 6 v t -»• 0- r t -> 0+ so h cannot be extended to a continuous function at a = 0. X — ij. X 9(8) 5 cos 9 h(t) *W = d + I/I) 1 ". fl = Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 114 Chapter 2 Limits and Continuity 36. From the graph we see that lim k(x) ^ lim k(x) x -> x -> 0+ so k cannot be extended to a continuous function at a = 0. k x) k(x) = 1 - 2W ' a = 37. (a) f(— 1) = — 1 and f(2) = 5 => f has a root between — 1 and 2 by the Intermediate Value Theorem, (b), (c) root is 1.32471795724 38. (a) f(— 2) = —2 and f(0) = 2 =>■ f has a root between —2 and by the Intermediate Value Theorem, (b), (c) root is -1.76929235424 CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES 1. (a) X 0.1 0.01 0.001 0.0001 0.00001 x x 0.7943 0.9550 0.9931 0.9991 0.9999 Apparently, lim x" x — » 0+ (b) 0.6 0.2 y = x 0.2 0.6 1 2. (a) x l/(l„x) C) 1,,,n ' 10 100 1000 0.3679 0.3679 0.3679 •i\VM Apparently, ^Hm, (±) A ° XJ = 0.3678 = 1 (b) 0.4. 0.2 f(x) J \ l/('» ») 6) 2 4 6 Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Chapter 2 Additional and Advanced Exercises 115 3. lim_ L V — » c lim_ Ln V — > c 1- lim v 2 l - o The left-hand limit was needed because the function L is undefined if v > c (the rocket cannot move faster than the speed of light). 1 < 0.2 -0.2 < 1 < 0.2 => 0.8 < Y < l - 2 =>■ L6 < V x < 2A => 2 - 56 < x < 5 - 76 - 1 <0.1 -0.1 < 1 < 0.1 => 0.9 < Y < L1 => 1.8 < a/x < 2.2 => 3.24 < x < 4.84. 5. 1 10 + (t - 70) x 10~ 4 - 10| < 0.0005 => |(t - 70) x 10~ 4 | < 0.0005 =>■ -5 < t - 70 < 5 => 65° < t < 75° => Within 5° F. -0.0005 < (t - 70) x 10~ 4 < 0.0005 1010 6. We want to know in what interval to hold values of h to make V satisfy the inequality |V - 1000 1 = |367rh - 1000| < 10. To find out, we solve the inequality: |367rh - 1000| < 10 =^ -10 < 367rh - 1000 < 10 => 990 < 367rh < 1010 =^> 8.8 < h < 8.9. where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe. The interval in which we should hold h is about 8.9 — 8.8 = 0.1 cm wide (1 mm). With stripes 1 mm wide, we can expect to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking. ^ < h < 367T — — 367T 7. Show lim f(x) = lim (x 2 - 7) = -6 = f(l). x — > 1 x — > 1 Step 1: |(x 2 - 7) + 6| < e => -e < x 2 - 1 < e =>■ 1 - e < x 2 < 1 + e => \J \ - t < x < a/I + e. Step 2: |x - 1| < 6 => -6 < x - 1 < 6 =$■ -6 + 1<x<6 + 1. Then -6+1= \/l - e or 6 + 1 = y/l + e. Choose 6 = min j 1 - \/ \ - e, y/l + e - l| , then < |x — ll < <5 =>• j(x 2 — 7) — 61 < e and lim f(x) = —6. By the continuity test, f(x) is continuous at x = 1. x — > 1 Show lim i g(x) = lim ^ = 2 4 X^ | x Step 1: | i - 2| < e =4> -e < ^ - 2 < e =4> 2 - e < ^ < 2 + e Step 2: |x- j| < 6 => -6 < x- \ < 6 =4> -6+\<x<6+\ 6 = — = — - — or 6 + - = — L - v 4 4+2e 4(2+e)' u1u ^4 4-2e Then -6 + \ 4 + 2e 4-2e ^ x ^ 4+2e ' Choose 6 = ,,„ £ , , , the smaller of the two values. Then < |x — t| < <5 => \w — 2| < e and lim ^- = 2. 4(z+e) 14 1 I zx I i 2x 4-2e I _]_ I 2x 4(2 -0 By the continuity test, g(x) is continuous at x 9. Show lim h(x) = lim J2x - 3 = 1 = h(2). x^2 x^ 2 v Stepl: I y/2x - 3 - ll < e =S> -e < V2X -3-1 < e =^> l-e< ^2x - 3 < 1 + e =5> (1 ~f +3 <x< (1 + f + 3 . Step 2: |x - 2| < 6 => -6<x-2<6oi-6 + 2<x<6 + 2. Then -5 + 2 = ^#±^ => 6 = 2 - ( ^4 ±A = ^^ = e - £ , or 8 + 2 (l+e) 2 + 3 2 s= (A±f+A_ 2 (i + f) 2 -i 2 e + |- . Choose <5 = e — y, the smaller of the two values . Then, < |x — 2| < 6 => v 2x — 3 — 1 < e, so lim \J 2x — 3=1. By the continuity test, h(x) is continuous at x = 2. 10. Show lim F(x) = lim ^9 - x = 2 = F(5). x — > 5 x — » 5 Step 1: l-x/9 — x — 2| < e =4> -e < ^9 - x - 2 < e =4> 9 - (2 - e) 2 > x > 9 - (2 + e) 2 Step 2: < |x - 5| < <*> =>• -<5 < x - 5 < <5 => -< < > + 5<x<<5 + 5. Then —(5 9-(2 + e) 2 =► <S = (2 + e) 2 2e, or * 9 - (2 - e) 2 =>• 6 = 4 - (2 - e) 2 = e 2 2e. Cfiojt (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 116 Chapter 2 Limits and Continuity Choose S 2e, the smaller of the two values. Then, < |x — 51 < 6 Vs 2 < e, so lim v 9 — x = 2. By the continuity test, F(x) is continuous at x = 5. x — > 5 11. Suppose Li and L 2 are two different limits. Without loss of generality assume L 2 > Li. Let e = I (L 2 — Li). Since lim f(x) = Li there is a <5i > such that < |x - x | < Si => |f(x) - Li| < e => -e < f(x) - Li < e X > Xq => - i (L 2 - Li) + Li < f(x) < 1 (L 2 - Li) + Li => 4Li - L 2 < 3f(x) < 2Li + L 2 . Likewise, Hm f(x) = L 2 so there is a 62 such that < |x — xo| < 82 => |f(x) — L 2 | < e => — e < f(x) — L 2 < e =► - i (L 2 - U) + L 2 < f(x) < i (L 2 - L : ) + L 2 => 2L 2 + L± < 3f(x) < 4L 2 - U =4> Li — 4L 2 < — 3f(x) < — 2L 2 — Lj. If 5 = min {61, 6 2 } both inequalities must hold for < |x — x | < S: 4Lj - L 2 < 3f(x) < 2Lj U - 4L 2 < -3f(x) < -2L 2 a contradiction. a + u \ L 2 - Li J => 5(Li - L 2 ) < < L x - L 2 . That is, Li - L? < and U - L 2 > 0, 12. Suppose lim f(x) = L. If k = 0, then lim kf(x) = lim = = 0- lim f(x) and we are done. cr X — > C X— > C X — > C X — > c If k ^ 0, then given any e > 0, there is a 6 > so that < |x - c| < 6 =>• |f(x) - L| < j|r =4> |k||f(x) - L| < e =$► |k(f(x) - L)| < e => |(kf(x)) - (kL)| < e. Thus, lim kf(x) = kL = k( lim f(x)Y 13. (a) Since x — > + , < x 3 < x < 1 =4> (x 3 - x) — ► 0" =4> lim f (x 3 - x) = lim f(y) = B where y = x 3 - x. v ; x-o+ y-o- : ' J (b) Since x — *• 0~, -1 < x < x 3 < => (x 3 - x) -^ 0+ =>• lim f (x 3 - x) = lim f(y) = A where y = x 3 - x. V ' X^O" V ' y ^0+ (c) Since x -*• 0+, < x 4 < x 2 < 1 =>■ (x 2 - x 4 ) -> 0+ => lim f(x 2 -x 4 )= lim f(y) = A where y = x 2 - x 4 . x — » + y — > 0+ (d) Since x -> 0~, -1 < x < => < x 4 < x 2 < 1 => (x 2 - x 4 ) -> 0+ =>■ lim f (x 2 - x 4 ) = A as in part (c). x — > 0+ 14. (a) True, because if lim (f(x) + g(x)) exists then lim (f(x) + g(x)) — lim f(x) = lim [(f(x) + g(x)) — f(x)] x — ► 2l x — ^ 3. x — y a x — * a = lim g(x) exists, contrary to assumption. (b) False; for example take f(x) = 1 and g(x) = — -. Then neither lim f(x) nor lim g(x) exists, but x ' x x^O x — > lim (f(x) + g(x)) = lim (I - i) = lim = exists, x — > x ^ x x x^O (c) True, because g(x) = |x| is continuous => g(f(x)) = |f(x)| is continuous (it is the composite of continuous functions). x < (d) False; for example let f(x) = < ' continuous at x = 0. x> f(x) is discontinuous at x = 0. However |f(x)| = 1 is x z -l 15. Show lim f(x) = lim , x — > — 1 x —> -1 x + ' x->-l ( x+1 > Define the continuous extension of f(x) as F(x) = < l im (» + ix» : D = _2, x ^-1, x -1 x ■+ _ 1 x + 1 ' ' . We now prove the limit of f(x) as x — » 1 -2 , x = — 1 exists and has the correct value. Step 1: x+l -(-2) < e => -e< (x+l)(x-l) (x+l) 2 < e => -e < (x - 1) + 2 < e, x^-1 =>• -e-l<x<e-l. Step 2: |x - (-1)| < (5 => -6 < x + 1 < S => -S-l<x<S-l. Then -6 - 1 = -e - 1 => S = e,OT 8 - I = e - 1 => 6 = e. Choose S = e. Then < |x - (-1)| < S \x l -l (—2) < e =$■ lim F(x) = —2. Since the conditions of the continuity test are met by F(x), then f(x) has a I x+1 'I x-> -1 continuous extension to F(x) at x = — 1 Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle Chapter 2 Additional and Advanced Exercises 117 16. Show lim g(x) = lim x — > 3 x — > 3 x 2 - 2x - 3 2x-6 lim x^ 3 (x-3)(x + l) Define the continuous extension of g(x) as G(x) 2(x-3) - 2, X 7^ 3. r x^-2x-3 x ^ 3 3 exists and has the correct value. x 2 -2x-3 2 < e => -e < (x - 3)(x + 1) Step 1: | 2x6 ^| ^ ^ -, >, - 2(x-3) Step 2: |x - 3| < 6 => -6<x-3<6 =>• 3-6 <x<6 2x - 6 ' ' . We now prove the limit of g(x) as 2< e => _ e< *±!_ 2 <e, x ^ 3 => 3 - 2e < x < 3 + 2e. Then, 3 - 6 = 3 - 2e I x 2 - 2x - 3 2x-6 2 < e =► lim ( *- ( f ^ x^3 2(x-3) 2e, or 6 + 3 = 3 + 2e => 5 = 2e. Choose <5 = 2e. Then < |x - 3| < 6 2. Since the conditions of the continuity test hold for G(x), g(x) can be continuously extended to G(x) at x = 3. 17. (a) Let e > be given. If x is rational, then f(x) = x => |f(x) — Oj = |x — 0| < e o- |x — 0| < e; i.e., choose 6 = e. Then |x — 0| < 6 =4> |f(x) - 0| < e for x rational. If x is irrational, then f(x) = => |f(x) - 0| < e ■& < e which is true no matter how close irrational x is to 0, so again we can choose 6 = e. In either case, given e > there is a 6 = e > such that < |x — 0| < <5 =>■ |f(x) — 0| < e. Therefore, f is continuous at x = 0. (b) Choose x = c > 0. Then within any interval (c — 6, c + 6) there are both rational and irrational numbers. If c is rational, pick e = |. No matter how small we choose 6 > there is an irrational number x in (c — 6, c + 6) =^ |f(x) — f(c)| = |0 — c| = c > | = e. That is, f is not continuous at any rational c > 0. On the other hand, suppose c is irrational => f(c) = 0. Again pick e = | . No matter how small we choose 6 > 3c -. Then |f(x) - f(c)| = |x - 0| e ■& | < x < there is a rational number x in (c — 6, c + 6) with |x — c| < | = = |x| > | = £ =>• f is not continuous at any irrational c > 0. If x = c < 0, repeat the argument picking e = '-y = ^ ■ Therefore f fails to be continuous at any nonzero value x = c. 18. (a) Let c = ™ be a rational number in [0, 1] reduced to lowest terms =^> f(c) = j. Pick e = ^. No matter how small 6 > is taken, there is an irrational number x in the interval (c — 6. c + 6) =>■ |f(x) — f(c)| = |0 — - 1 - > tj- = e. Therefore f is discontinuous at x = c, a rational number. n 2n ' (b) Now suppose c is an irrational number =>• f(c) = 0. Let e > be given. Notice that | is the only rational number reduced to lowest terms with denominator 2 and belonging to [0, 1]; I and | the only rationals with denominator 3 belonging to [0, 1]; \ and | with denominator 4 in [0, 1]; \, |, I and I with denominator 5 in [0, 1]; etc. In general, choose N so that i < e =^> there exist only finitely many rationals in [0, 1] having denominator < N, say ri, r2, ..., r p . Let 6 = min{|c — r ; | : i=l,...,p}. Then the interval (c — 6. c + 6) contains no rational numbers with denominator < N. Thus, < |x — c| < 6 =^> |f(x) — f(c)| = |f(x) — 0| = |f(x)| < i < e => f is continuous at x = c irrational. Copyright (c) 2006 Pearson Education 118 Chapter 2 Limits and Continuity (c) The graph looks like the markings on a typical ruler when the points (x, f(x)) on the graph of f(x) are connected to the x-axis with vertical lines. 0.8 0.6 0.4 0.2" m. im m 0.2 0.4 0.6 0.8 f( , = f 1 /" itx = m/ni /w [0 if x is irratio; is a rational number in lowest terms irrational 19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero point, 0, on the equator =4> + irR represents the midnight point (at the same exact time). Suppose Xj is a point on the equator "just after" noon =4> Xj + irR is simultaneously "just after" midnight. It seems reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after midnight: That is, T(xi) — T(xi + 7rR) > 0. At exactly the same moment in time pick x-2 to be a point just before midnight => X2 + irR is just before noon. Then T(X2) — T(X2 + ttR) < 0. Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c between (noon) and ttR (simultaneously midnight) such that T(c) — T(c + 7rR) = 0; i.e., there is always a pair of antipodal points on the earth's equator where the temperatures are the same. 20. x lim f(x)g(x) = x lim 1 [(f(x) + g(x)) 2 - (f(x) - g(x)) 2 ] = \ [( x lim (f(x) + g(x)))' - ( x lim(f(x) - g(x)) H3 2 -(-ir 21. (a) Atx = 0: lim r + (a) = lim a — > a — > -1 + Jl + i lim a^0 lim l-(l+a) a ^0 af-1-Vl + a) -l-yi + O At x = — 1: lim r + (a) = lim a — > — 1 + a — (b) At x = 0: lim r_ (a) = lim a — > a — » l-(l + a) ! )fe£§) lim 1+ a(-l-Vl + a) a->-l a (-1-^/1+ a) -1 - v^ 1- x/TTe a^0" V a / \-l + s/T+IJ lim t-q + a) lim lim -l a^0" a(-l + Vl + a) a -> 0" a (-1 + ^1 + a) a _, Q" -1 + Vl + a oo (because the denominator is always negative); lim r_(a) = lim -, = — oo (because the denominator a^0+ - 1 + 1 is always positive). Therefore, lim r_ (a) does not exist. a — > At x = — 1: lim r_(a) a->-l+ lim -1- Vl + a lim 4— = 1 _> _1+ -l + V!+ a Copyright (c) 2006 Pearson Education Chapter 2 Additional and Advanced Exercises 119 (c) (d) -0.5 0.5 1 Graph not to scale r (a) a=0.05 1 1 / r-(o)=- -\--JTTa a 1 -2 2 ^_4___ S- 8 -4 a=0.2 22. f(x) = x + 2 cos x =>• f(0) = + 2 cos = 2 > and f(-7r) = ~n + 2 cos (-tt) = -n - 2 < 0. Since f(x) is continuous on [— ir, 0], by the Intermediate Value Theorem, f(x) must take on every value between [—tt — 2, 2]. Thus there is some number c in [— tt, 0] such that f(c) = 0; i.e., c is a solution to x + 2 cos x = 0. 23. (a) The function f is bounded on D if f(x) > M and f(x) < N for all x in D. This means M < f(x) < N for all x in D. Choose B to be max {|M| , |N|} . Then |f(x)| < B. On the other hand, if |f(x)| < B, then B < f(x) < B => f(x) > -B and f(x) < B =>■ f(x) is bounded on D with N = B an upper bound and M = — B a lower bound. (b) Assume f(x) < N for all x and that L > N. Let e L-N Since lim f(x) = L there is a 6 > such that xn < |x - x | < 6 => |f(x) - L| < e <S> L - e < f(x)< L + e <S> L - ^ < f(x)< L + ^ O ^±S < f(x) < 3L-N But L > N L + N f(x) < N. This contradiction proves L < N. > N => N < f(x) contrary to the boundedness assumption (c) Assume M < f(x) for all x and that L < M. Let e = ^y^. As in part (b), < |x - x j < 6 < f(x) < L + <£• 3L-M < f(x) < < M, a contradiction. 24. (a) If a > b, then a - b > => |a - b| = a - b => max (a, b) = ^ i + b I a — b _ 2 ' 2 — 2 If a < b, then a - b < =>• = ^ =b 2 u - (b) Let min (a, b) - ^ - ^^ (a — b) = b — a =4> max (a, b) a + b 2 |a-b| 2a 2 a + b 2 b-a 2 a + b 2 Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 120 Chapter 2 Limits and Continuity 25. lim sin(l — cos x) sin(l-cosx) 1 -cosx 1 + cos x : lim lim sin(l — cos x) lim )r cos2x , = 1 ■ lim 1 + cos x x j, q 1 — cos x x ^ q x(l + cos x) x j, q x(l + cos x) lim ^ . ^mjL_ =1 . (2) =0 . x _^ q x 1 + cos x V 2 / 26. lim -^- = lim ^ • -^ %. x _» o+ sm V x x -» + x Sln V x V x 1 • lim t-^y • lim v/x = 1 • • = 0. 27. lim ™(^) = lim *M^A . m* = i im ^i^l . lim ^ = , . i = L x^O x^O sin x x x^O x^O 28. lim sin(x 2 + x) _ j- m sin(x 2 +x) . ( v j_ i\ _ ]; m sin(x 2 + x) x^O x x ^o x2 + x (x + 1) = lim =J^2 • lim (x + 1) = 1 • 1 = 1 V ; x^O x2 + x x^0 V 29. lim x^2 sin(x 2 - 4) _ ,. sin(x 2 - 4) sin(x 2 — 4) , - lim ^f^l . ( x + 2) = lim sn ^=^ • lim (x + 2) = 1 • 4 = 4 x - 2 x^2 x2 - 4 V ; x^2 x2 - 4 x^2 V ' 30. lim ^(^-3) = lim Mf-z) . _^ = lim Mf-z) . , im x^9 x ~ 9 - = 1 • - = - x't^'g v^- 3 \/ x + 3 x-T9 \A- 3 x i ^' 1 9\/ x + 3 6 6 Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle CHAPTER 3 DIFFERENTIATION 3.1 THE DERIVATIVE OF A FUNCTION 1. Stepl: f(x) = 4-x 2 andf(x + h) = 4-(x + h) 2 Stet) 2' f ( x + h >- f W = K-(x + h) 2 ]-(4-x 2 ) _ (4-x 2 -2xh-h 2 )-4 + x 2 _ -2xh-h 2 _ h(-2x-h) " ' h h h h h = -2x - h Step 3: f (x) = lim (-2x - h) = -2x; f (-3) = 6, f (0) = 0, f (1) = -2 h — > 2. F(x) = (x - l) 2 + 1 and F(x + h) = (x + h - l) 2 + 1 => F'(x) = lim [(x + h-l) 2 + l]-[(x-l) 2 +l] h-»0 ,■ (x 2 + 2xh + h 2 -2x-2h+l + l)-(x 2 -2x+l + l) _ h™0 h 2(x - 1); F'(-l) = -4, F'(0) = -2, F'(2) = 2 lim 2xh + !f- 2h = lim (2x + h - 2) h->0 h->0 3. Stepl: g(t)= iandg(t + h) Step 2: i _ i . g(t + h)-g(t) _ (t + h)2 t2 (t + h) 2 / l 2 -(l + h) 2 h h h(-2t-h) _ -2t-h (t + h) 2 t 2 h (t + h) 2 1 2 t 2 -(t 2 + 2th + h 2 ) _ -2th - h 2 (t + h) 2 -t 2 -h (t + h) 2 t 2 h Step 3: g'(t) = h lim o ^^ = $ = f ; g'(-l) = 2, g'(2) = - ±, g' ( ^3 | 2 3x/3 4. k(z) = ^ and k(z + h) = ^^> => k'(z) = lim a-(z+h) _ i_z > , 2(z + h) 2z J h->0 lim h-»0 lim h->0 2(z + h)zh h^To 2(z + h)zh ^:k'i-l) = -i,k'(l) = -i,k'(v^)=-i lim lim h — > o 2 < z + h ' zh h — > 2 ' z + h ' 5. Stepl: p(0) = ^/Je and p(6> + h) = ^3(0 + h) Step 2- P( fl + h >-P< fl > = v^cfl + h)->/3fl = (x^Tsh-v^) _ (y/3flT3h+y3e) = (3 e + 3h)-3e h h h (v/3fl + 3h+x/3fl) h(v/30 + 3h+y3e) 3h ^VM+lh+V^fl) V'3ff + 3h+ v /3fl Step 3: p'(0) = lim . 3 — ^ = ^ ^- - F FV y h->0 x/39 + 3h + y/ie s/36+s/3fJ 2x739 — -P'(l)= A,p'(3) = i,p'^ 2V3 2'*" V3^ 2n /2 6. r(s) = ^/2^+T and r(s + h) = v /2(sTh)~+T => r'(s) = lim 72s + 2h+ 1 - x/2s+ 1 lim h-»0 (v^s+h+l- V2s + l) (x/2s + 2h+l + v^s+l) h->0 lim (2s + 2h+l)-(2s+l) (V'2s + 2h+l + v /2s+l) h^O h( v /2s + 2h+l + v'2s + l) lim 2h lim h( v /2l+2h+T+v / 2sTT) h-»0 \72s + 2h+l + v/2s+l x/2s + 1 + 72s + 1 2^25 + 1 73 i 7T ;r'(0)=l,r'(l)=^,r'(i) = ^ t c / \ i 3 j f/ i u\ i/i u\3 > dy i- 2(x + h) 3 -2x 3 ,- 2 (x 3 + 3x 2 h + 3xh 2 + h 3 ) - 2x 3 7. y = f(x) = 2x° and f(x + h) = 2(x + hr =>■ / = lim — r = lim — r dx h->0 h h->0 h lim 6x 2 h + 6xh 2 + 2h 3 = Hm dx h(6x 2 + 6xh + 2h 2 ) h-»0 h->0 lim (6x 2 + 6xh + 2h 2 ) = 6x 2 h-»0 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 122 Chapter 3 Differentiation dr lim h->0 H^ + l -[f + l] 1 lim [(s + h) 3 + 2]-[s 3 + 2] h^O lim h->0 s 3 + 3s 2 h + 3sh 2 + h 3 + 2-s 3 -2 _ 1 j im h[3s 2 + 3sh + h 2 ] _ l , ; . vi ; , ,„■ ,,.,., , ,.,_., 3 ,'J h->0 lim (3s 2 + 3sh + h 2 ) = § s 2 h->0 9. s = r(t) = ^-r and r(t + h) t+h 2(t+h)+l ds dt lim h->0 (,2(t + h)+lJ (21+1) lim h->0 lim h->0 ' (t + h)(2t+l)-t(2t + 2h+l) N , (2t + 2h+l)(2t+l) 1 lim (t + h)(2t+l)-t(2t + 2h+l) 1 2t 2 + 1 + 2ht + h - 2t 2 - 2ht - t (2t + 2h+l)(2t+l)h _ 1 q (2t + 2h+l)(2t+l)h lim h _ .,.,, IfZTd (2t + 2h+l)(2t+l)h h^To (2t + 2h + l)(2t+l) lim (2t+l)(2t+l) (2t+l) 2 10. dv lim h->0 »+•»-!& -('- lim ht 2 + h 2 t + h lim — = lim — ^ h->0 h t 2 + ht + 1 _ t 2 + 1 _ lim h-»0 ' h(t + h)t-t + (t + hA v (t + h)t j h" h (' + h > t h" (' + ••)« 11. p = f(q) = -^i- and f(q + h) v^q + hjTT dp _ jj m W(q + h) + i7 l^q+T/ d 1 h -> h lim h->0 >/q + 1 - \/i + '' + 1 \/q + h+l ^q+1 lim /q+l - y/q + h+l h h ^ h^q + h+1 Vq+1 lim (v/q^-y/q + h + i) . (\/q+7+\/q+h+T) = lim (q+l)-(q + h+l) h->[) h ^q + h + 1 ^/q + 1 (^q+l + ^/q+h+l) h ^ h^q + h+l ^/q + 1 (^q + 1 + ,/q + h + l) lim -h lim h^o h x / q +h+i % /q + i(\/q + 1 + \/q + h + 1 ) h->o ^q+h+i v / q + i( v /q+i + Vq+ h + 1 ) ^i -1 yq + 1 yq - 1 ( s/q + 1 - \/q + ' ) 2 (q+ ') \Ai- ' 12. ^ = lim dw ^3(w + h)-2 ^3w-2J _ j im ^3w - 2 - V3w + 3h - 2 h^O h h^O h v /3w + 3h-2 v / 3w-2 (v/3w-2- v/3w + 3h-2~) (' % /3w-2+ v /3w+3h-2~) h-»0 h v /3w + 3h-2V3w-2 (^/3w-2 + y/3w + 3h-2\ lim (3w-2)-(3w + 3h-2) h->0 h v/3w + 3h - 2 ./3w - 2 (y3w - 2 + ./3w + 3h - 2) lim -3 I) !) v/3w + 3h - 2 \/3w - 2 (V 3w - 2 + ^/3w + 3h - 2) v/3w - 2 ^/3w - 2 (y3w - 2 + \/3w - 2) -3 2(3w - 2) V3W-2 13. f(x) = x + 2 and f(x + h) = (x + h) f(x + h)-f(x) _ [(" + h )+(7Th)] -[" + !] x(x + h) 2 + 9x - x 2 (x + h) - 9(x + h) _ x 3 + 2x 2 h + xh 2 + 9x - x 3 - x 2 h - 9x - 9h _ x 2 h + xh 2 -9h x(x + h)h x(x + h)h x(x + h)h h(x 2 + xh-9) _ .^±xj^9 ;f , (x)= ^ 4±^_9 = S 2 _9 = 1 _ 9 ;m = f( _ 3) = x(x + h)h x(x + h) h ^0 x < x + h > 14. k(x) = ^ and k(x + h) = ^^ lim (2 + x) - (2 + x + h) lim k'(x)= lim k(x + h >- k(x) h->0 h lim lim h->0 ^2+x+h 2+ h) h" h(2 + x)(2 + x + h) h x ^' h(2 + x)(2 + x + h) h ^"g (2 + x)(2 + x + h) (2 + x)'- k'(2) 16 15. ds lim h->0 [(t + h) 3 -(t + h) 2 ]-(t 3 -t 2 ) lim h->0 (t 3 + 3t 2 h + 3th 2 + h 3 ) - (t 2 + 2th + h 2 ) - 1 3 + 1 2 lim h->0 3t 2 h + 3th 2 + h 3 -2th-h 2 Hm h(3t 2 + 3 t h + h 2 -2 t -h) = nm (3t2 3th h2 _ 2t _ h) h^O h h->0 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 3.1 The Derivative of a Function 123 3t 2 - 2t; m : ds I dt I i=-l 16- g lim h->0 (x + h+l) 3 -(x+l) 3 lim h->0 (x + l) 3 + 3(x + l) 2 h + 3(x + l)h 2 + h 3 -(x + l) 3 lim [3(x + l) 2 + 3(x + l)h + h 2 ] = 3(x + l) 2 ; m h-»0 dy dx 17. f(x) and f(x + h) V(x + h)-2 8 (\/x-2- v/x + h-2) (y/x-2 + Vx + h-2) f(x + h) - f(x) _ y/|x + h)-2 ^~2 h ~~ h 8[(x-2)-(x + h-2)] h v /x + h-2 v / x-2 (Vx-2 + Vx + h-2) hv/x + h - 2 V x - 2 ( \/x - 2 + \A + h - 2 ) -8h h\/x + h-2\/x-2(v'x-2 + \/x + h-2j -8 -4 f'(x)= lim r 8 =■, h^O \/x + h - 2 Vx - 2 (yx - 2 + y/x + h - 2 J - ; m = f (6) = jtj = — \ =>■ the equation of the tangent Vx - 2 ^x - 2 ( Vx - 2 + Vx - 2 j (x-2)Vx line at (6, 4) is y - 4 = - ± (x - 6) ^ y = - |x + 3 + 4 ^ y = - ±x + 7 ( 1 + ^4 - (z + h)) - f 1 + \A^z) 18. g'(z) = lim - — ^ — i = lim 6 w h->0 h h->0 \£ lim (4 - z - h) - (4 - z) lim 4-z-h-\/4-z lim [V4-z-h+v4-zJ (V 4 - z - h + ./4^) h^O h(V4-z-h+\/4-z) h^O h (^4- z-h+ ^4- z) h^O (\/4-z - h+ ^4- z) m = g'(3) 2V4-3 i => the equation of the tangent line at (3, 2) is w — 2 = — \ (z — 3) w -Iz+f + 2=>w=-iz+|. 19. s = f(t) = 1 - 3t 2 and f(t + h) = 1 - 3(t + h) 2 = 1 - 3t 2 - 6th - 3h 2 => % = lim f(i + h)-f(t) h->0 lim h->0 (l-3t 2 -6th-3h 2 )-(l-3t 2 ) lim (-6t - 3h) = -6t — I =6 dt I t=-i ° 20. y = f(x) = 1 - I and f(x + h) = 1 - ^ => & =-. lim f(x + h , ) ~ f(x) = lim dx 1 L_ lim x , x+h = lim , ''.,. = lim -A^x h->0 h h->0 "(x + h)h h ^ x(x + h) h^O h-»0 dy dx -Si 21. r = f(0) and f(0 + h) lim '4-1 2\/4-8-2\/4-e-h x/4-(9 + h) dr _. j im f(9 + h)-f(9) _ lim y-4-fl-h ^9 h->0 h->0 _ ini| 2^4^8-2^4^^ (2\/4~^ + 2V4~g-h) ^""u hv/4^9v/4-9-h h'Tu hy/4^6y/4-B-h ' (2V^fl + 2V4-e-h) lim lim 4(4 - 9) - 4(4 - 9 - h) lim h^O 2h ^4 - 9 ^4 - 8 - h ( ^4 - 6 + y/4 - 8 - h) h^O \/4-0 ^4-8 -h(i/4-8 + \/4-8 -h) 2 _ 1 . dr I _ 1 ~^ de\g=o 8 (4-9) (2^4-9) (4~9)V4- 22. w = f(z) = z + ypL and f(z + h) = (z + h) + y/z + h (z + h + v/z + h) - (z + yz) dz lim h->0 f(z + h) - f(z) lim 1 + lim (z + h)-z h-»0 h[Jz + h- ■y/i) lim h->0 1 + lim h+ Jz + h-Jz h->0 Vz + h+^z lim h->0 2,/z dw I 5 dz I z=4 4 /z + h + ,/z -h+v/z") 23. f'(x) = lim f(z) ~ f(x) lim z — > x z-x z^x z-x lim (x + 2)-(z + 2) lim lim z^»x(z-x)(z + 2)(x + 2) z ^»x(z-x)(z + 2)(x + 2) z " x(z + 2)(x + 2) (x + 2)^ Copffigl (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle 124 Chapter 3 Differentiation _j 1 fi2 (\_n2 24. f (x) = lim fi^I^iW = Hm fr=ff Eg = i im (»-i)Mz-i)' = Hm v ' z^x z-x z — ► x z-x z — > x(z-x z-ir x- 1)- z^x [{x . 1) . {z . lMx . 1) + {z . 1)] 2/ ,%2 lim (x — z)(x + z — 2) lim (z-x)(z-inx-l)- z^x (z-x)(z-l)^(x-l) l(x + z-2) _ -l(2x-2) _ -2(x-l) _ -2 z^x(z-x)(z-l)"(x-l)^ z^x(z-l)"(x-ir (x-1) 4 (x-1) 4 (x-1) 25. g'(x) = lim g(z) - g(x) lim z(x-l)-x(z-l) 1 j JQ z-1 »-l _ llln __ |1|N _ |u|1 z^x z-x z —> x z-x z — > x(z-x)(z- l)(x- 1) z — > x(z — x)(z-l)(x-l) z — > x(z- l)(x- 1) lim lim (x-ir 26. g'(x) = lim g(z) ' g(x) z^x z-x z^x z-x lim <i +> A)-(i+^ = lim ^/w^ . a^ = lim — z-x ' ' z— >x z-x ^z + ^x z — » x(z-x)(,/z + v/x) = lim r L t- = ^V z — » x s/z + ^/x 2^/x 27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x = 0), then positive =S> the slope is always increasing which matches (b). 28. Note that the slope of the tangent line is never negative. For x negative, f^(x) is positive but decreasing as x increases. When x = 0, the slope of the tangent line to x is 0. For x > 0, f^(x) is positive and increasing. This graph matches (a). 29. fs(x) is an oscillating function like the cosine. Everywhere that the graph of f3 has a horizontal tangent we expect fg to be zero, and (d) matches this condition. 30. The graph matches with (c). 31. (a) f is not defined at x = 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. slope of line joining (—4, 0) and (0, 2) = I but lim For example, lim fW ~^ (0) F x -> o- x - ° line joining (0, 2) and (1, —2) = —4. Since these values are not equal, f'(0) = lim f(x) - f(0) x^0 (b) > > . 3 _ /'on (-4, 6) 2 o-o , , <^-A 'o 1 6 ' ' > -8-6-4-2 _ 2 4 6 8 r x^0 + x-0 f(x) - f(0) x-0 slope of does not exist. 32. (a) (b) Shift the graph in (a) down 3 units y Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 3.1 The Derivative of a Function 125 33. 84 ,«, 87 -3.3 - 0-<i 34. (a) dp/dt 20 x 16 n S 8 t (days] 10 20 30 40 50 (b) The fastest is between the 20 lh and 30" 1 days; slowest is between the 40 lh and 50 lh days. 35. Left-hand derivative: For h < 0, f(0 + h) = f(h) = h (using y = x curve) =4> lim_ h — > lim h->(T h 2 -0 lim h = 0; twO" Right-hand derivative: For h > 0, f(0 + h) = f(h) = h (using y = x curve) lim h^O 4 h-0 lim 1 = 1; h->0+ f(0 + h)-f(0) f(0 + h) - f(0) lim lwO+ f(0 + h) -f(0) h f(0 + h) - f(0) Then Inn !lif ' 4i = " la + , lim^ " U + "^ 1W => the derivative f'(0) does not exist. h->(r h->0+ 36. Left-hand derivative: When h < 0, 1 + h < 1 => f(l+h) = 2 = lim = 0; h-»rr h — (T h lim h->(T Right-hand derivative: When h > 0, 1 + h > 1 => f(l + h) = 2(1 + h) = 2 + 2h => lim h — > + f(l+h)-f(l) h lim Then lim h-»(T £±|^ = Um a = lim 2 = 2 h h^0+ h h->0 + f(l+h)-f(l) j_ ,- m f(l+h)-f(l) ^ lim h->0 the derivative f'(l) does not exist. 37. Left-hand derivative: When h < 0, 1 + h < 1 => f(l + h) = x/l+h => lim f(1 + h . ) ~ f(1) v h -> 0" h lim IwO" '1 +h-l lim h->(T '1+h-lJ h lim (l+h)-l lim 1+h+l) h->(T hfVl+h + l) h->(T v/T+h+l 2 ' Right-hand derivative: When h > 0, 1 -I- h > 1 => f(l + h) = 2(1 + h) - 1 = 2h + 1 => lim f(l+h)-f(l) l im < 2h +')- 1 = lim 2 = 2; h->0+ h->0+ f(l+h)-f(l) _j_ lim f(l + h)-f(l) h-0+ h Then lim ^=pi^ =£ lim h -> o- h the derivative f'(l) does not exist. 38. Left-hand derivative: lim h->o _ Right-hand derivative: lim 5 h-0+ f(l+h)-f(l) f(l+h)-f(l) lim = lim h->0+ (l+h)-l lim 1 = 1; h->0~ lim - — j— - h^0+ h lim -h -1 lim -1; Then lim f(1 + h >- f(1) ± lim f(1 + h >- f(1) h-»(T h->(T the derivative f'(l) does not exist. Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 126 Chapter 3 Differentiation 39. (a) The function is differentiable on its domain — 3 < x < 2 (it is smooth) (b) none (c) none 40. (a) The function is differentiable on its domain — 2 < x < 3 (it is smooth) (b) none (c) none 41. (a) The function is differentiable on — 3 < x < and < x < 3 (b) none (c) The function is neither continuous nor differentiable at x = since lim f(x) ^ lim f(x) x -> x -» 0+ 42. (a) f is differentiable on -2 < x < -1, -1 < x < 0, < x < 2, and2 < x < 3 (b) f is continuous but not differentiable at x = — 1 : lim f(x) = exists but there is a corner at x X — » — 1 1 since lim f( - 1+h) - f '- 1) h->u" -3 and lim h->0+ f(-l + h)-f(-l) h h -► 0+ h (c) f is neither continuous nor differentiable at x = and x = 2: f'(— 1) does not exist at x = 0, lim f(x) = 3 but lim f(x) = = x -> 0" x -> 0+ at x = 2, lim f(x) exists but lim f(x) d f(2) x^2 x->2 w ^ v lim f(x) does not exist; x — > 43. (a) f is differentiable on -1 < x < and < x < 2 (b) f is continuous but not differentiable at x = 0: lim f(x) = exists but there is a cusp at x = 0, so x — ► f'(0) = lim f(0 + h)-f(0) h->0 does not exist (c) none 44. (a) f is differentiable on -3 < x < -2, -2 < x < 2, and 2 < x < 3 (b) f is continuous but not differentiable at x = —2 and x = 2: there are corners at those points (c) none 45. (a) f'(x) = ,lim f(x + tl ^ f(x) (b) h->0 y l lim h->0 -(x + h) 2 -(-x 2 ) h lim h->0 -x 2 — 2xh — h 2 + x 2 lim (— 2x — h) h->0 -2x y = -x -2x -1 (c) y' = — 2x is positive for x < 0, y' is zero when x = 0, y' is negative when x > (d) y = —x 2 is increasing for — oo < x < and decreasing for < x < oo; the function is increasing on intervals where y' > and decreasing on intervals where y' < 46. (a) f'(x) = lim f(x + h)-f(x) h->0 lim i -i -n i X+Jl x ; lim -x + (x + h) h — > x ( x + h ) h lim ; , , , = -4 h — » o x ( x + h ) x Copffigl (c) 1 Pearson %$A\ k, publishing as Pearson Addison-Wesle Section 3.1 The Derivative of a Function 127 (b) y = ■ (c) y' is positive for all x ^ 0, y' is never 0, y' is never negative (d) y = — - is increasing for — oo < x < and < x < oo 47. (a) Using the alternate formula for calculating derivatives: f'(x) = lim — — — = lim -^ - _ Hm (z-x)(z 2 + zx + x 2 ) = Um z — > x 3(z - x) z — > x 3(z - x) z — ► x lim z3 - x3 f'(x) = X 2 (b) (c) y' is positive for all x/0, and y' = when x = 0; y' is never negative (d) y = y is increasing for all x ^ (the graph is horizontal at x = 0) because y is increasing where y' > 0; y is never decreasing M - jjto 48. (a) Using the alternate form for calculating derivatives: f'(x) = lirn '""_ " " - - Jim z ^x ^ = m (z ' x)(z3 4 r- 2 xr 2z+x3 - -x 7 r y z^x z-x z^x z-x i = lim (*-*)(z 3 +Xz2+A+x3) = JJ Z 3 + XZ3 + A+X3 = X 3 ^ f'(x) = X 3 z — > x 4(z - x) z — > x 4(z - x) z-tl J (c) y' is positive for x > 0, y' is zero for x = 0, y' is negative for x < (d) y = x is increasing on < x < oo and decreasing on — oo < x < (x— c) (x 2 + xc + c 2 ) 49. y' = lim ^^^ = lim ^^ = lim "- CMX ' + xc + r) = lim (x 2 + xc + c 2 ) = 3c 2 J x^c x-c x^c x-c x^c x-c x ^ c v / The slope of the curve y = x 3 at x = c is y' = 3c 2 . Notice that 3c 2 > for all c =>• y = x 3 never has a negative slope. 2,/x 4- h — 2 /x 50. Horizontal tangents occur where y' = 0. Thus, y' = lim — — r — — b J J h->0 h = lim 2 (^^) (vgh+^) = lim ^((x + hj-x)) = Hm 2 i h->0 h (v/xTh+v/x"] h->0 h^Vx + h+v^J h -> \/x + h + \A V x Cop # (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 128 Chapter 3 Differentiation Then y' = when A= = which is never true => the curve has no horizontal tangents. 51. y' = lim 1 h->0 (2(x + h) 2 - 13(x + h) + 5) - (2x 2 - 13x + 5) h lim h->0 2x 2 + 4xh + 2h 2 - 13x - 13h + 5 - 2x 2 + 13x - 5 lim h->0 4xh + 2h 2 -13h lim (4x + 2h — 13) = 4x — 13, slope at x. The slope is —1 when 4x — 13 h->0 •1 > 4x = 12 => x = 3 => y = 2-3 2 — 13-3 + 5= 16. Thus the tangent line is y + 16 = (-l)(x - 3) y = — x — 13 and the point of tangency is (3, —16). 52. For the curve y = -Jx, we have y' lim h->0 (Vx + h- x/x) (\A + h +V / *) lim (x + h) - x (\A + h + yit) h^O (\/x + h+x/x) h lim , } — y= - h^O V x + h +V x 2^/x y- . Suppose (a, ^/a) is the point of tangency of such a line and (— 1, 0) is the point on the line where it crosses the x-axis. Then the slope of the line is ^zrz\\ = irj which must also equal . - tV => 2a = a + 1 => a=l. Thus such a line does a+ 1 2\/a t^ ; using the derivative formula at x = a exist: its point of tangency is (1, 1), its slope is — k= = ^; and an equation of the line is y — 1 = | (x — 1) ^y=ix+i. 53. No. Derivatives of functions have the intermediate value property. The function f(x) = |_xj satisfies f(0) = and f(l) = 1 but does not take on the value | anywhere in [0, 1] =>■ f does not have the intermediate value property. Thus f cannot be the derivative of any function on [0, 1] => f cannot be the derivative of any function on (—oo, oo). 54. The graphs are the same. So we know that for f(x) = |x| , we have f'(x) = — . 1 I -i <H 55. Yes; the derivative of — f is — f so that f'(xo) exists =>• — f'(xo) exists as well. 56. Yes; the derivative of 3g is 3g' so that g'(7) exists => 3g'(7) exists as well. 57. Yes, lim fi2 can exist but it need not equal zero. For example, let g(t) = mt and h(t) = t. Then g(0) = h(0) 0, but lim g(0 lim o w t->0 lim m = m, which need not be zero. t-»0 58. (a) Suppose |f(x)[ < x 2 for -1 < x < 1. Then |f(0)| < 2 => f(0) = 0. Then f'(0) = lim gO + h) - f(0) h^O lim h->0 f(h)-0 lim h->0 f(h) For |h| < 1, -h 2 < f(h) < h 2 =>■ -h < ^ < h => f'(0) = lim f(h) h->0 by the Sandwich Theorem for limits, (b) Note that for x ^ 0, |f(x)| = |x 2 sin A| = |x 2 | |sinx| < |x 2 | • 1 = x 2 (since -1 < sin x < 1). By part (a), f is differentiable at x = and f'(0) = 0. 59. The graphs are shown below for h = 1, 0.5, 0.1. The function y = =K- is the derivative of the function 2y x /x so that — 7- = lim V x h->0 The graphs reveal that y — — r — — gets closer to y = — K Cop # (c) 1 Pearson Education, to, publishing as Pearson Addison-Wesle Section 3.1 The Derivative of a Function 129 as h gets smaller and smaller. y h = 0.5 h = 0.1 60. The graphs are shown below for h = 2, 1, 0.5. The function y = 3x 2 is the derivative of the function y = x 3 so that 3x 2 = lim (x+h ^ x . The graphs reveal that y h-»0 h gets smaller and smaller. (x+h) 3 -x 3 gets closer to y = 3x 2 as h 61. Weierstrass's nowhere differentiable continuous function. y *M = cosfrx) + Qj cos(9jrjr) + (J\ cosi^nx) + f|Ycos(9V*) \j\ cos&irx) H h-O.ZU 2 // (x + h)< - X> k \ 1 4> -10 12 + ••• + 62-61. Example CAS commands: Maple : f := x -> x A 3 + x A 2 - x; Copyright (c) 1 Pearson E Won, k, publishing as Pearson Addison-Wesle 130 Chapter 3 Differentiation x0:= 1; plot( f(x), x=xO-5..xO+2, coloi-black, title="Section 3_1, #62(a)" ); q := unapply( (f(x+h)-f(x))/h, (x,h) ); # (b) L := limit( q(x,h), h=0 ); # (c) m := eval( L, x=x0 ); tan_line := f(x0) + m*(x-x0); plot( [f(x),tan_line], x=x0-2..x0+3, color=black, linestyle=[l,7], title="Section 3.1 #62(d)", legend=["y=f(x)", "Tangent line at x=l"] ); Xvals := sort( [ xO+2 A (-k) $ k=0..5, xO-2 A (-k) $ k=0..5 ] ): # (e) Yvals := map( f, Xvals ): evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >); plot( L, x=x0-5..x0+3, color=black, title="Section 3.1 #62(f)" ); Mathematica : (functions and xO may vary) (see section 2.5 re. RealOnly ): «Miscellaneous"RealOnly" Clear[f, m, x, y, h] x0= 7T /4; f[x_]:=x 2 Cos[x] Plot[f[x], {x,x0 - 3,x0 + 3}] q[x_,h_]:=(f[x + h]-f[x])/h m[x_]:=Limit[q[x,h], h ->• 0] ytan:=f[x0] + m[x0] (x - xO) Plot[{f[x], ytan},{x, xO - 3, xO + 3}] m[x0 - 1]//N m[x0 + 1]//N Plot[{f[x], m[x]},{x, xO - 3, xO + 3}] 3.2 DIFFERENTIATION RULES 1. y = -x 2 + 3 => dy - A f_ x 2^ i A dx dx V A ) ~ dx (-x 2 ) + £ (3) = -2x + = -2x =>• d 2 y 2. y = x 2 + x + 8 => a = 2x+l+0 = 2x+l =>• d 2 v dx- 1 3. s = 5t 3 - 3t 5 =► | = | (5t 3 ) - g (3t 5 ) = 15t 2 - 15t 4 =* § = | (15t 2 ) - | (15t 4 ) = 30t - 60t 3 4. w = 3z 7 - 7z 3 + 21z 2 =>■ g = 21z° - 21z 2 + 42z =>■ ff = 126z 5 - 42z + 42 4 v 3 _ v . .. dy _ 4v 2 _ , ^ dfy = gx 5. y = | x 3 - x ^ ^y = 4x 2 - 1 ^ 6. y=^ + ^ + | =>. d| =x 2 +x+ l ^ g=2x+l+0 = 2x+l 7. w = 3z~ 2 -z" 1 => 1 _, dw -6z" -6 | 1 ^ = 18z- 4 -2z- 3 =if-| 8. s = -2T 1 + 4t- 2 => | = 2t- 2 - 8r 3 = | - | =* |§ = -4t- 3 + 24r 4 = ^ + £ 9. y = 6x 2 - lOx - 5x~ 2 => jjz = I2x - 10 + 10x~ 3 = 12x - 10 + ^ => f$ = 12 - - 30x~ 4 = 12 - 2° Cfiojt (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 3.2 Differentiation Rules 131 10. y = 4 - 2x - x- 3 => ^ -2 + 3x~ 4 = -2+3^g=0- 12x- 5 = -F 11. r= I s -2_I s -l dr ds 2 E -3 i 5 „-2 _ ^2 | _5_ 3 s t 2 5 ~~ 3s 3 "^ 2s 2 d^r d>- 2s~ 4 - 5s- 3 = 4 - t 12. r = 126I- 1 -46>- 3 +6>- 4 24 48 | 20 | = -l26>- 2 + 126> 4 46» -*> - rJ2 _i_ 12 _ £ " 92 T 04 05 |§ = 246»- 3 - 48<r 5 + 206>" 6 13. (a) y = (3 - x 2 ) (x 3 - x + 1) => y' = (3 - x 2 ) • £ (x 3 - x + 1) + (x 3 - x + 1) - £ (3 - x 2 ) = (3 - x 2 ) (3x 2 - 1) + (x 3 - x + 1) (-2x) = -5x 4 + 12x 2 - 2x - 3 (b) y = -x 5 + 4x 3 - x 2 - 3x + 3 => y 1 = -5x 4 + 12x 2 - 2x - 3 14. (a) y = (x - 1) (x 2 + x + 1) => y' = (x - l)(2x + 1) + (x 2 + x + 1) (1) = 3x 2 (b) y = (x - 1) (x 2 + x + 1) = x 3 - 1 => y' = 3x 2 15. (a) y = (x 2 + l)(x + 5+i) => y' = (x 2 + l)-£(x + 5+i) + (x + 5 + i)-£(x 2 + l) = (x 2 + 1) (1 - x- 2 ) + (x + 5 + x- 1 ) (2x) = (x 2 - 1 + 1 - x- 2 ) + (2x 2 + lOx + 2) = 3x 2 + lOx + 2 - X 2 (b) y = x 3 + 5x 2 + 2x + 5 + i =>■ y' = 3x 2 + lOx + 2 - 4j 16. y=(x+I)(x-I + l) (a) y' = (x + x- 1 ) - (1 + x- 2 ) + (x - x" 1 + 1) (1 - x~ 2 ) = 2x + 1 - A + | (b) y = x 2 + x + 1 - A. => y' = 2x + 1 - i + x 2 + x 3 17. y = f^l ; use the quotient rule: u = 2x + 5 and v = 3x - 2 =>■ u' = 2 and v' = 3 =$■ y' = V "'~ 2 UV ' _ (3x-2)(2)-(2x + 5)(3) _ 6x-4-6x-15 _ -19 (3x - 2) 2 (3x-2) 2 (3x-2) 2 18. z 2x+l . dz _ (x 2 -l)(2)-(2x+l)(2x) _ 2x 2 - 2 - 4x 2 - 2x _ -2x 2 - 2x - 2 _ -2(x 2 + x+l) x 2 — 1 dx (x 2 -l) 2 (x 2 -l) 2 (x 2 -l) 2 (" 2 -ir 19. g(x) = |tq| ; use the quotient rule: u = x 2 — 4 and v = x + 0.5 => u' = 2x and v' = 1 =>■ g'(x) _ (x + 0.5)(2x)-(x 2 -4)(l) _ 2x 2 + x-x 2 + 4 _ x 2 + x + 4 (x + 0.5) 2 (x + 0.5) 2 (x + 0.5) 2 t 2 -l _ (t-l)(t + l) _ t+1 t 2 + t-2 ~~ (t- - U - 1!! » - ,. i i- > - Ct+2)(t-l) ~~ t+2' L V^ 1 =* I (V) t(t\ - (t + 2)(l)-(t+l)(l) = t + 2-t-l = 1 (t + 2) 2 (t + 2) 2 (t + 2) 2 21. V = (l-t)(l+t 2 r 1 = i^ => dx = (l+t 2 )(-l)-0-'>(2t) _ -l- t 2 -2t + 2t 2 _ t^jt^l 22. w x + 5 2x-7 (1+t 2 ) 2 ~ (1+t 2 ) 2 (1+t 2 ) 2 I _ (2x-7)(l)-(x + 5)(2) _ 2x - 7 - 2x - 10 _ -17 (2x - 7) 2 (2x-7) 2 (2x-7) 2 23. f(s) vA-i f'(s) (y^ 1 )^)-^-')^) _ (^+i)-(^-i) v^+i " ' '" ~ (Vi+i) 2 2^(^+1)" ^(%A+i) 2 NOTE: ^ (,/s) = ^V from Example 2 in Section 2.1 24. u 5x+l 2~7x du dx (2 v ^)(5)-(5x+l)(-ij) 4x 5x- 1 4x 3 / 2 Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 132 Chapter 3 Differentiation 25. v 1 +x-4Jx £ x(l-^)-(l+x-V3 _ 2 ^_ 26. r = 2 ITS of v^o)-i(^) ! 1 a 12 27. y = ( x 2_ 1 wx 2 + x+n > use tne quotient rule: u = 1 and v = (x 2 — 1) (x 2 + x + 1) => u' = and v' = (x 2 - 1) (2x + 1) + (x 2 + x + 1) (2x) = 2x 3 + x 2 - 2x - 1 + 2x 3 + 2x 2 + 2x = 4x 3 + 3x 2 - 1 dy vu' — uv' dx v 2 0-l(4x 3 + 3x 2 -l) -4x 3 - 3x 2 + 1 (x 2 - l) 2 (x 2 + x+ 1)" (x 2 - If (x 2 + x+ 1)" 28. y (x+l)(x + 2) x 2 + 3x + 2 . / (x 2 - 3x + 2) (2x + 3) - (x 2 + 3x + 2) (2x - 3) _ -6x 2 + 12 (x-l)(x-2) x 2 -3x + 2 = ^ J (x-l) 2 (x-2) 2 (x-l) 2 (x-2) 2 -6 (x 2 - 2) (x-l) 2 (x-2) 2 29. y 30. y 31. y 32. S: 1 „4 3 v 2 x 2 - x => y' = 2x 3 - 3x - 1 => y" = 6x 2 - 3 => y'" = 12x =4> y (4) = 12 =>• yM = for all n > 5 i-x 5 =^ y' 120 x J +7 24 A v " _ I v 3 _^ ..III _ 1 2 y — , a => y — „ a /(4) =^ y( 5 ' = 1 => yM = for all n > 6 x 2 + 7X- 1 =► p- = 2x - 7x- 2 = 2x - \ dx x 2 g=2+14x- 3 = 2+i| ^±|M = 1 + f - P = 1 + St" 1 - r 2 =► |=0- 5r 2 + 2r 3 = -5r 2 + 2r 3 = ^ + | d 2 ! _ -, n t -3 _ fit -4 _ 10 _ 6 dt 2 _ lul Dt - 7 P 33. r: (g-D(fl 2 + g+i) 3 W 3 3-3 ^ = + 3(T 4 = 3<T 4 A , eft e» ^ do 2 -i2r 5 = ^ 34. u (x 2 + x)(x 2 -x+l) _ x(x+l)(x 2 -x+l) _ x(x 3 + l) _ x " + > 1 + A = 1 + x -3 1=0- 3x- 4 -3x- 4 = =| => & = 12x- 5 = if x 4 dx 2 x J 35. w ;i±^) (3 - z) = (i z- 1 + 1) (3 - z) = z- 1 - i + 3 - z = z- 1 dw 0-1 d-\v dz-' 2z- 3 - = 2z -3 _ 2 36. w = (z + l)(z - 1) (z 2 + 1) = (z 2 - 1) (z 2 + 1) = z 4 - 1 => f- = 4z 3 - = 4z 3 => d 2 w dz 2 12z 2 37. p 38. p I q 2 + 3 \ / q 4 - 1 \ _ q 6 - q 2 + 3q 4 V i2q / V i 3 / 12 i 4 d 2 P _ 1 1 „-4 -3 1 „2 1 „-2 ill „-4 . dp _ 1 , 1 3 , 5 _ 1 q '-si dq 6 q + i q + q" 6q 3 dq 2 — 6 2 1 q 2 + 3 5q -6 _ 1 1 ' 6 2q 4 q 2 + 3 q 2 + 3 _ q 2 + 3 J_ - I a -l (q-l) 3 + (q+l) 3 ~~ (q 3 -3q 2 + 3q-l) + (q 3 + 3q 2 + 3q+l) — 2q 3 + 6q — 2q (q 2 + 3) — 2q — 2 4 dp dq iq- 2 .J_ =s. ^E _ -3 _ J_ 2q 2 ^ dq 2 — 4 — q 3 39. u(0) = 5, u'(0) = -3, v(0) = -1, v'(0) = 2 (a) £ (uv) = uv' + vu' =* £ (uv)| x=o = u(0)v'(0) + v(0)u'(0) = 5 - 2 + (-l)(-3) = 13 d (a\ _ vu'-uv' . d (u\\ _ v(0)u'(0) - u(0)v / (0) _ (-l)(-3) - (5)(2) _ n (b) (C) dx \vJ v 2 ~^ dx VvVlx-o (v(0)) 2 (-1) 2 _d_ (v\ _ uv' - vu' . _d_ fv\| _ u(0)v'(0) - v(0V(0) _ (5)(2)-(-l)(-3) _ _7_ dx I u / — u 2 ^ dx V u ) I x =o ~~ (u(0)) 2 ~~ (5) 2 ~~ 25 Copffigl (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle (d) £ (7v - 2u) = 7v' - 2u' =>• £(7v-2u)| Section 3.2 Differentiation Rules 133 7v'(0) - 2u'(0) = 7 • 2 - 2(-3) = 20 40. u(l) = 2, u'(l) (a) (b) (c) (d) s(«v)| dx dx \uJ 0, v(l) = 5, v'(l) = - u(l)v'(l) + v(l)u'(l) v(l)u'(l)-u(l)v'(l) 2-(-l) + 5-0 (V(l))2 u(l)v'(l)-v(l)u'(l) (U(l))2 5-0— 2-(— 1 ) (5)2 2-(-l)-5-0 (2)2 2_ 25 dx (7v - 2u)| _ = 7v'(l) - 2u'(l) = 7 - (-1) - 2 - = -7 41. y = x 3 - 4x + 1. Note that (2, 1) is on the curve: 1 = 2 3 - 4(2) + 1 (a) Slope of the tangent at (x, y) is y' = 3x 2 - 4 => slope of the tangent at (2, 1) is y'(2) = 3(2) 2 -4 = 8. Thus the slope of the line perpendicular to the tangent at (2, 1) is — | =>• the equation of the line perpendicular to to the tangent line at (2, 1) is y — 1 = — | (x — 2) or y = — f + f • (b) The slope of the curve at x is m = 3x 2 — 4 and the smallest value for m is —4 when x = and y = 1. (c) We want the slope of the curve to be 8 => y' = 8 => 3x 2 — 4 = 8 = x = 2, y = 1 and the tangent line has equation y — 1 = 8(x — 2) or y y = (— 2) 3 — 4(— 2) +1 = 1, and the tangent line has equation y — 1 : 3x 2 = 12 =^> x 2 = 8x — 15; when x = 8(x + 2) or y = 8x 4 =>• x -2, 1-17. ±2. When 42. (a) y x 3 — 3x — 2 =4> y' = 3x 2 — 3. For the tangent to be horizontal, we need m = y' = => = 3x 2 — 3 =>• 3x 2 = 3 =4> x = ±1. When x = — 1, y = => the tangent line has equation y = 0. The line perpendicular to this line at (— 1, 0) is x = — 1. When x = 1, y = — 4 =>• the tangent line has equation y = —4. The line perpendicular to this line at (1, —4) is x = 1. (b) The smallest value of y' is —3, and this occurs when x = and y = —2. The tangent to the curve at (0, —2) has slope —3 =>■ the line perpendicular to the tangent at (0, —2) has slope i y y- 2 = i (x - 0) or 43. y 4x It. 3 * - 2 is an equation of the perpendicular line. =>• dy dx _ (x 2 +l)(4)-(4x)(2x) (x2+l) 2 4x 2 + 4-8x 2 (x2+l)2 . 4(- (x x 2 + l + 1) 2 X 2 +l = 4, so the tangent to the curve at (0, 0) is the line y = 4x. When x curve at (1, 2) is the line y = 2. 4(0+1) 1 When x = 0, y = and y' 1 , y = 2 => y' = 0, so the tangent to the 44. (x 2 + 4)(0)-8(2x) -16x When x = 2, y = 1 and y' ) x 2 + 4 -r J ~ (x 2 + 4) 2 (x 2 + 4) 2 line to the curve at (2, 1) has the equation y — 1 = — I (x — 2), or y -16(2) (2 2 + 4) 2 1-2. 5 , so the tangent 45. y = ax 2 + bx + c passes through (0, 0) =>■ = a(0) + b(0) + c 2 = a + b; y' = 2ax + b and since the curve is tangent to y = x at the origin, its slope is 1 at x = y' = 1 when x = => 1 = 2a(0) + b =>• b = 1. Then a + b = 2 =>- a=l. In summary a = b - 0; y = ax 2 + bx passes through (1, 2) 1 and c = so the curve is y v-2 46. passes through (1, 0) => = c(l) — 1 1 — 2x and x = 1 — 1. Since y = x — x 2 and y 1 =>• the curve is y = x — x . For this curve, x 2 + ax + b have common tangents at x = 0, x 2 + ax + b must also have slope — 1 at x = 1 . Thus y' = 2x + a =>• — 1 = 2-1 y = x 2 — 3x + b. Since this last curve passes through (1, 0), we have = 1 — 34 —3, b = 2 and c = 1 so the curves are y = x 2 — 3x + 2 and y = x — x 2 . a => a = — 3 3 =>• b = 2. In summary, 47. (a) y y x° - x =>• y 2(x + 1) or y --3x 2 - 1. Whenx = -l,y 2x + 2. and y' = 2 =>• the tangent line to the curve at (—1, 0) is Copyright (c) 2006 Pearson Education 134 Chapter 3 Differentiation (b) (c) * * I => x 3 - x = 2x + 2 =>■ x 3 - 3x - 2 = (x - 2)(x + l) 2 = =>• x = 2 or x = - 1. Since y = 2(2) + 2 = 6; the other intersection point is (2, 6) 48. (a) y = x 3 - 6x 2 + 5x => y' = 3x 2 - 12x + 5. When x = 0, y = and y' = 5 (0, 0) is y = 5x. (b) the tangent line to the curve at (c) y = x 3 — 6x 2 + 5x y = 5x Since y = 5(6) = 30, the other intersection point is (6, 30). 6x 2 + 5x = 5x =^ x 3 - 6x 2 = =^ x 2 (x - 6) = =4> x = 0orx = 6. 49. P(x) = a n x n + a n _!X n a 2 x 2 + ajx + a => P'(x) = na n x n : + (n — l)a n _ 1 x l n-2 + 2a 2 x + ai 50. R = M 2 (f - f ) = f M 2 - | M 3 , where C is a constant dR CM-M 2 51. Let c be a constant =>■ gf = => |^(u-c) = u-^ + c-^=u-0 + c^ = c^. Thus when one of the functions is a constant, the Product Rule is just the Constant Multiple Rule =>■ the Constant Multiple Rule is a special case of the Product Rule. n I dv ■< dv 52. (a) We use the Quotient rule to derive the Reciprocal Rule (with u=l): f- (-) = T JkL = — 2^ v 2 dx »• A. (^ (b) Now, using the Reciprocal Rule and the Product Rule, we'll derive the Quotient Rule. K v N ; d\ = u • E (v) + v • I (Product Rule) = u-(^)^ + ^ (Reciprocal Rule) *-f u .V dx V v / v 2 = dx y2 dx , the Quotient Rule. 53. (a) A (uvw)= A ((uv) . w) = (uv) dw +w d (uv) = uv dw +w(u d^ +v ^ =uv dw +wu d^ +wv d^ dx uvw' + uv'w + u'vw (b) £ (U1U2U3U4) = £ ((U1U2U3) U 4 ) = (U!U 2 U 3 ) |j± + U 4 £ (U1U2U3) U lU2 U 3 £ + U 4 ( Ul U 2 £ + U 3Ul £ + U 3 U 2 £; dxV— »/ - S (U 1U2 U 3 U 4 ) (using (a) above) =* A (u lU2 u 3 u 4 ) = u lU2 u 3 to + Ul u 2 u 4 to + Ul u 3 u 4 to + u 2 u 3 u 4 to = UiU 2 u 3 u 4 + UiU 2 u 3 u 4 + uiu 2 u 3 u 4 + uiu 2 u 3 u 4 (c) Generalizing (a) and (b) above, 4- (ur • -u n ) = Uiu 2 - • -u^iu^ + Uiu 2 - • ■u„_ 2 u' 1 _ 1 u 11 + ... + u'^- ■ -u n Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 3.3 The Derivative as a Rate of Change 135 54. In this problem we don't know the Power Rule works with fractional powers so we can't use it. Remember d (\/*) = TT~ (f rom Example 2 in Section 2.1) dx (a) aH* 3/2 ) = aH* • * 1/2 ) = * • £ (\A) + (b) ^s« = x -2t; 'x-l + /^ = !^ = 3 x l/2 ( X 5/2) = A ( x 2 . x l/2) = x 2 ^ ^ + ^ _d_ (x 2) = x 2 . ^_±_j + ^ ■ 2x = ± X 3 / 2 + 2x 3 / 2 = § X 3 / 2 ( c ) £ ( x7/2 ) = £ ( x3 • xV2 ) = x3 £ (v^) + V 7 ^ £ ( x3 ) = x3 • (jfe) + \A • 3x2 = 3 x5/2 + 3x5/2 = \ x5/2 (d) We have £ (x 3 / 2 ) = | x 1 / 2 , £ (x 5 / 2 ) = f x 3 / 2 , £ (x 7 / 2 ) = | x 5 / 2 so it appears that £ (x 11 / 2 ) = § x^ 2 )" 1 whenever n is an odd positive integer > 3. 55. p nRT dP dV V- We are holding T constant, and a, b, n, R are also constant so their derivatives are zero -nRT (V-nb)-0-(nRT)(l) _ V 2 (0) - (an 2 ) (2V) (V-nb) 2 (y2) 2 ~~ (V-nb) 2 2iin- V 3 56. A(q) = s + cm hq 2 (km)q- 1 +cm+ (|)q=> dA dq -(km)q- 2 +(|) km I h ' q 2 + 2 dt 2 2(km)q -3 _ 2km - q3 3.3 THE DERIVATIVE AS A RATE OF CHANGE t 2 - 3t + 2, < t < 2 As At (a) displacement = As = s(2) — s(0) = Om — 2m = —2 m, v !iv - v — (b) v = | = 2t - 3 => |v(0)| = |-3| = 3 m/sec and |v(2)| = 1 m/sec; a = ff = 2 => a(0) = 2 m/sec 2 and a(2) = 2 m/sec 2 -1 m/sec (c) v = 0=>2t — 3 = => t = |. v is negative in the interval < t < | and v is positive when | < t < 2 =>■ the body changes direction at t s = 6t - t 2 , < t < 6 (a) displacement = As = s(6) — s(0) = m, v. lv = 4| = | = m/sec (b) v = | = 6 - 2t => |v(0)| = | 6| = 6 m/sec and |v(6)| = |-6| = 6 m/sec; a = |f = -2 => a(0) = -2 m/sec 2 and a(6) = -2 m/sec 2 (c) v = => 6 — 2t = =4> t = 3. v is positive in the interval < t < 3 and v is negative when 3 < t < 6 => the body changes direction at t = 3. d£s dt 2 -6t + 6 s = -t 3 + 3t 2 - 3t, < t < 3 (a) displacement = As = s(3) — s(0) = —9 m, v av = ^| = ^ = — 3 m/sec (b) v = | = -3t 2 + 6t - 3 => |v(0)| = |-3| = 3 m/sec and |v(3)| = |-12| = 12 m/sec; a => a(0) = 6 m/sec 2 and a(3) = — 12 m/sec 2 (c) v = => -3t 2 + 6t - 3 = => t 2 - 2t + 1 = =^> (t - l) 2 = => t = 1. For all other values of t in the interval the velocity v is negative (the graph of v = — 3t 2 + 6t — 3 is a parabola with vertex at t = 1 which opens downward => the body never changes direction). 4. s f 3 i a t 2 , < t < 3 (a) As = s(3) - s(0) m, v.,. As At m/sec (b) v = t 3 - 3t 2 + 2t => |v(0)| = m/sec and |v(3)| = 6 m/sec; a = 3t 2 - 6t + 2 => a(0) = 2 m/sec 2 and a(3) = 11 m/sec 2 (c) v = => t 3 - 3t 2 + 2t = => t(t - 2)(t - 1) = => t = 0, 1, 2 => v = t(t - 2)(t - 1) is positive in the interval for < t < 1 and v is negative for 1 < t < 2 and v is positive for 2 < t < 3 => the body changes direction at Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 136 Chapter 3 Differentiation t = 1 and at t = 2. 5. s= f - 5 1 < t<5 v- t ' — — (a) As = s(5) - s(l) = -20 m, v a -20 4 -5 m/sec 150 10 t 4 ,3 a(l) = 140 m/sec 2 and (b) v = ^|° + | => |v(l)| = 45 m/sec and |v(5)| = \ m/sec; a a(5) = j~ m/sec 2 (c) v = =4> ~ 5 ° 3 +5t = =4> -50 + 5t = =>• t = 10 => the body does not change direction in the interval 6. s= ^, -4 < t < (a) As = s(0) - s(-4) = -20 m, v av 20 -5 m/sec (b) v -25 (t+5) 2 l v (~ 4)| = 25 m/sec and |v(0)| = 1 m/sec; a 50 (t+5) 3 a(— 4) = 50 m/sec 2 and a(0) = % m/sec 2 (c) v = => 25 )2 =0 =>- v is never => the body never changes direction s = t 3 — 6t 2 + 9t and let the positive direction be to the right on the s-axis. (a) v = 3t 2 - 12t + 9 so that v = =>■ t 2 -4t + 3 = (t- 3)(t - 1) = =>• t = 1 or 3; a = 6t - 12 => a(l) = —6 m/sec 2 and a(3) = 6 m/sec 2 . Thus the body is motionless but being accelerated left when t = 1, and motionless but being accelerated right when t = 3. (b) a = =>■ 6t - 12 = => t = 2 with speed | v(2)| = 1 12 - 24 + 9| = 3 m/sec (c) The body moves to the right or forward on < t < 1, and to the left or backward on 1 < t < 2. The positions are s(0) = 0, s(l) = 4 and s(2) = 2 =4> total distance = |s(l) - s(0)| + |s(2) - s(l)| = |4| + |-2| = 6 m. v = t 2 -4t + 3 =4> a = 2t-4 (a) v = => t 2 - 4t + 3 = =>• t = 1 or 3 => a(l) = -2 m/sec 2 and a(3) = 2 m/sec 2 (b) v > =>■ (t - 3)(t - 1) > =>■ < t < 1 or t > 3 and the body is moving forward; v < =*> (t - 3)(t - 1) < => 1 < t < 3 and the body is moving backward (c) velocity increasing =4> a>0 => 2t-4>0 => t>2; velocity decreasing =4> a<0 =>• 2t-4<0 => < t < 2 1.86t 2 3.72t and solving 3.72t = 27.8 => t « 7.5 sec on Mars; s i = 11.44t 2 22.88t and solving 22.88t = 27.8 =^ t w 1.2 sec on Jupiter. 10. (a) v(t) = s'(t) = 24 - 1.6t m/sec, and a(t) = v'(t) = s"(t) = -1.6 m/sec 2 (b) Solve v(t) = => 24 - 1.6t = => t = 15 sec (c) s(15) = 24(15) -. 8(15) 2 = 180m (d) Solve s(t) = 90 => 24t - .8t 2 =90 => t = 30± ' 2 5v ^ w 4.39 sec going up and 25.6 sec going down (e) Twice the time it took to reach its highest point or 30 sec 11. s = 15t - \ g s t 2 => v = 15 - g s t so that v = => 15 - g s t = => g s = l -f ■ Therefore g s = ^ = | = 0.75 m/sec 2 12. Solving s m = 832t - 2.6t 2 = => t(832 - 2.6t) = => t = or 320 => 320 sec on the moon; solving s e = 832t - 16t 2 = =^ t(832 - 16t) = => t = or 52 =^> 52 sec on the earth. Also, v m = 832 - 5.2t = => t= 160 and s m (160) = 66,560 ft, the height it reaches above the moon's surface; v e = 832 — 32t = =>• t = 26 and s e (26) = 10,816 ft, the height it reaches above the earth's surface. 13. (a) s = 179 - 16t 2 -32t => speed = |v| = 32t ft/sec and a -32 ft/sec 2 Copyright (c) 2006 Pearson Education Section 3.3 The Derivative as a Rate of Change 137 (b) s = =$> 179 - 16t 2 = =>■ t (c) Whent 179 -32 179 "if « 3.3 sec -8\/l79« -107.0 ft/sec 14. (a) lim v = lim 9.8(sin 9)t = 9.8t so we expect v = 9.8t m/sec in free fall "2 "2 (b) a = f t = 9.8 m/sec 2 15. (a) at 2 and 7 seconds (c) \v\ (m/sec) (b) between 3 and 6 seconds: 3 < t < 6 (d) Speed ■ ' ' V i i i i V i i \ , , (sec) 2 4 6 8 10 4 - dv 3 — o <^-^ 2 , _ 1 23456789 10 -2 o^— -3 -<^—o -4 16. (a) P is moving to the left when 2<t<3or5<t<6;Pis moving to the right when < t < 1; P is standing still when l<t<2or3<t<5 (b) (cm/sec) i r 2 velocity -o— O — O o t (sec) 12 3 5 6 speed (cm/sec) 4 -O— o — o o 12 3 5 6 t (sec) 17. (a) 190 ft/sec (c) at 8 sec, ft/sec (e) From t = 8 until t = 10.8 sec, a total of 2.8 sec (f) Greatest acceleration happens 2 sec after launch (g) From t = 2 to t = 10.8 sec; during this period, a = v( '° 8 8 ) :^ (2) (b) 2 sec (d) 10.8 sec, 90 ft/sec -32 ft/sec 2 18. (a) Forward: < t < 1 and 5 < t < 7; Backward: 1 < t < 5; Speeds up: 1 < t < 2 and 5 < t < 6; Slows down: < t < 1, 3 < t < 5, and 6 < t < 7 (b) Positive: 3 < t < 6; negative: < t < 2 and 6 < t < 7; zero: 2<t<3and7<t<9 (c) t = and 2 < t < 3 (d) 7 < t < 9 19. s = 490t 2 v = 980t => a = 980 s(4/7)-s(0) 4/7 280 cm/sec. (a) Solving 160 = 490t 2 => t = I sec. The average velocity was (b) At the 160 cm mark the balls are falling at v(4/7) = 560 cm/sec. The acceleration at the 160 cm mark was 980 cm/sec 2 . (c) The light was flashing at a rate of jk = 29.75 flashes per second. Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 138 Chapter 3 Differentiation 20. (a) (b) 100 50 -50 -100 -150 -200 -50 a 20 a«30-6t 10 12 14 21. C = position, A = velocity, and B = acceleration. Neither A nor C can be the derivative of B because B's derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position. Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That leaves B for acceleration. 22. C = position, B = velocity, and A = acceleration. Curve C cannot be the derivative of either A or B because C has only negative values while both A and B have some positive slopes. So, C represents position. Curve C has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is negative where B has negative slopes and positive where B has positive slopes. 23. (a) c(100) = 11,000 11,000 100 $110 (b) c(x) = 2000 + lOOx - .lx 2 =^ c'(x) = 100 - .2x. Marginal cost = c'(x) =4> the marginal cost of producing 100 machines is c'( 100) = $80 (c) The cost of producing the 101 s1 machine is c(101) - c(100) = 100 - ^ = $79.90 24. (a) r(x) = 20000(1 - ±) (b) r'(100) = ?M» = $2. r'(x) 20000 which is marginal revenue. (c) lim r'(x) lim ;o(iiio 0. The increase in revenue as the number of items increases without bound will approach zero. 25. b(t) = 10 G + 10 4 t - 10 3 t 2 => b'(t) = 10 4 - (2) (10 3 t) = 10 3 (10 - 2t) (a) b'(0) = 10 4 bacteria/fir (b) b'(5) = bacteria/hr (c) b'(10) = -10 4 bacteria/hr 26. Q(t) = 200(30 - t) 2 = 200 (900 - 60t + t 2 ) => Q'(t) = 200(-60 + 2t) => Q'(10) = -8,000 gallons/min is the rate the water is running at the end of 10 min. Then Q j Q( = — 10,000 gallons/min is the average rate the water flows during the first 10 min. The negative signs indicate water is leaving the tank. Copyright (c) 2006 Pearson Education Section 3.3 The Derivative as a Rate of Change 139 27. (a) y = 6(l-^) 2 = 6(l-i + ^) =* $ = £-1 (b) The largest value of ^ is m/h when t = 12 and the fluid level is falling the slowest at that time. The smallest value of ^ is — 1 m/h, when t = 0, and the fluid level is falling the fastest at that time. the graph of y is dy (c) In this situation, ^ < always decreasing. As -^ increases in value, the slope of the graph of y increases from — 1 to over the interval < t < 12. y S 5 4 3 y- 6(,- \2> 1 ^ "12 "dy~ t dt 12 28. (a) V = | Trr 3 =>• ^ = 4vrr 2 => ^| r _ 2 = 4tt(2) 2 = 16tt ft 3 /ft (b) When r = 2, ^ = 167T so that when r changes by 1 unit, we expect V to change by approximately \6ir. Therefore when r changes by 0.2 units V changes by approximately (167r)(0.2) = 3.2tt w 10.05 ft 3 . Note that V(2.2) - V(2) w 11.09 ft 3 . 29. 200km/hr: t = 25, D = 55 ^m/sec i?(25) 2 500 10 t 2 6250 4 m/sec, and D= y t m ft. ThusV 500 9 20 500 9 t = 25 sec. When 30. s = v t - 16t 2 =>- v = v - 32t; v = =>• t = f§ ; 1900 = v t ^ v = v/^OCl^OO) = 80^ ft/sec and, finally, ^vj[ft 60 sec 1 min 16r so that t 60 min 1 hr 1 mi 5280 ft => 1900 238 mph. 32 64 31. , 600 ^ s = 20te-16r 2 400 200 / 4^ = 200 - / 1 l^*"M -32< \ -200 s 1 - ^=-32 d? >V "--^ 12 body moves down (a) v = when t = 6.25 sec (b) v > when < t < 6.25 =>■ body moves up; v < when 6.25 < t < 12.5 (c) body changes direction at t = 6.25 sec (d) body speeds up on (6.25, 12.5] and slows down on [0, 6.25) (e) The body is moving fastest at the endpoints t = and t = 12.5 when it is traveling 200 ft/sec. It's moving slowest at t = 6.25 when the speed is 0. (f) When t = 6.25 the body is s = 625 m from the origin and farthest away. Copyright (c) 2006 Pearson Education 140 Chapter 3 Differentiation 32. 33. (a) v = when t = | sec (b) v < when < t < 1.5 =^ body moves down; v > when 1.5 < t < 5 =4> body moves up (c) body changes direction at t = | sec (d) body speeds up on (|, 5] and slows down on [0, |) (e) body is moving fastest at t = 5 when the speed = |v(5)| = 7 units/sec; it is moving slowest at t = | when the speed is (f) When t = 5 the body is s = 12 units from the origin and farthest away. (a) v = when t = ^f^- sec (b) v < when ^/^ < t < ^of^ => body moves left; v > when < t < ^/^ or ^f^- < t < 4 => body moves right (c) body changes direction at t = — f- — sec (d) body speeds up on (^^, 2) U (^±^> 4 1 and slows down on lo, ^^) U (2, ^4^) . (e) The body is moving fastest at t = and t = 4 when it is moving 7 units/sec and slowest at t = — f — se< (f) When t = ^j- — the body is at position s « —6.303 units and farthest from the origin. Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 3.4 Derivatives of Trigonometric Functions 141 34. 10 /V\ 5 / s =4 - 7t -5 / 1 /\ 2\ 3 \ 4 dt / + 12t- 3t 2 \ -10 d J . .... A + 6t 2 -t 3 - t —5 - 1 2 - 6t (a) v = when t = ^f^ (b) v < when < t < 6 ~f^ or 6 + f^ < t < 4 =4> body is moving left; v > when 6-^/Tj < ( < 6+^As ^ body is moving right (c) body changes direction at t = — f- — sec (d) body speeds up on ( — f- — , 2 I U ( — ^ — 5 4 1 and slows down on |0, 6-y/H )u( 2 , 6 ±#) (e) The body is moving fastest at 7 units/sec when t = and t = 4; it is moving slowest and stationary at 6±-v/l5 t 3 (f) When t = -^f- — the position is s w 10.303 units and the body is farthest from the origin. 35. (a) It takes 135 seconds. AF (b) Average speed = ^- 5 - 73-0 yij « 0.068 furlongs/sec. (c) Using a symmetric difference quotient, the horse's speed is approximately 4f - ,,,.. .,.. - — 4-2 0.077 furlongs/sec. (d) The horse is running the fastest during the last furlong (between the 9th and 10th furlong markers). This furlong takes only 1 1 seconds to run, which is the least amount of time for a furlong. (e) The horse accelerates the fastest during the first furlong (between markers and 1). 3.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 1. y = — lOx + 3 cos x 2. y = - + 5 sin x 3. y = esc x — 4,/x + 7 =>• 10 + 3 -f (cosx) = -10 -3 sin x - - # + 5£(sinx)= ^+5cosx dx -CSC x cot x — -A- + = —csc x cot X 2,/x 4. y = x 2 cot x ^ dx -x 2 csc 2 x + 2x cot x x 2 £ (cot x) + cot x • i (x 2 ) + 4j = -x 2 csc 2 x + (cot x)(2x) (sec x + tan x) ^ (sec x — tan x) + (sec x — tan x) -r- (sec x + tan x) 5. y = (sec x + tan x)(sec x — tan x) => -j| - ,..^ . .„,, , w Jn = (sec x + tan x) (sec x tan x — sec 2 x) + (sec x — tan x) (sec x tan x + sec 2 x) = (sec 2 x tan x + sec x tan 2 x — sec 3 x — sec 2 x tan x) + (sec 2 x tan x — sec x tan 2 x + sec 3 x — tan x sec 2 x) = ( Note also that y = sec 2 x - tan 2 x = (tan 2 x + 1 ) - tan 2 x = 1 =4> ^ = 0. J Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 142 Chapter 3 Differentiation 6. y = (sin x + cos x) sec x => g| = (sin x + cos x) -4 (sec x) + sec x -g (sin x + cos x) (sin x + cos x) sin x . cos x - sin x dx v ' dx (sin x + cos x)(sec x tan x) + (sec x)(cos x — sin x) sin 2 x + cos x sin x + cos 2 x — cos x sin x 1 + sec - x ( Note also that y = sin x sec x + cos x sec x = tan x + 1 =>• i = sec 2 x. I 7 v — cotx . dy _ (1 + cot x ) s ( cot x ) ~ (cot x > s (' + cot x ) _ (1 + cot x) (-esc 2 x) - (cot x) (-esc 2 x) ' J ~ 1+cotx dx — (1+cotx) 2 (1+cotx) 2 —esc 2 x — CSC 2 x cot X + CSC 2 x cot x — CSC 2 X — (1+cotx) 2 (1 +cotx) 2 _ cosx . dy _ (1 + sin x) ^ (cos x) - (cos x) ^ (1 + sin x) _ (1 + sin x) (-sin x) - (cos x) (cos x) * 1+sinx dx (1+sinx) 2 (1+sinx) 2 — sin x — sin 2 x — cos 2 x — sin x — 1 —(1+sinx) —1 (1+sinx) 2 (1+sinx) 2 (1+sinx) 2 1+sinx 9. y = — — h — — = 4 sec x + cot x => -r = 4 sec x tan x — esc 2 x J cos x tan x dx 10. y + dy x(-sinx)-(cosx)(l) . (cos x)(l) - x(-sin x) _ -x sin x - cos x ■ cosx + x sin x x cos x dx x 2 cos 2 x x 2 cos 2 x 11. y = x 2 sin x + 2x cos x — 2 sin x =S> ■£ = (x 2 cos x + (sin x)(2x)) + ((2x)(— sin x) + (cos x)(2)) — 2 cos x = x 2 cos x + 2x sin x — 2x sin x + 2 cos x — 2 cos x = x 2 cos x 12. y = x 2 cos x — 2x sin x — 2 cos x =>• g| = (x 2 (— sin x) + (cos x)(2x)) — (2x cos x + (sin x)(2)) — 2(— sin x) = —x 2 sin x + 2x cos x — 2x cos x — 2 sin x + 2 sin x = — x 2 sin x 13. s = tan t - t =4> f t = | (tan t) - 1 = sec 2 1 - 1 = tan 2 t 14. s = t 2 - sec t + 1 =>• 21 = 2t - f (sec t) = 2t - sec t tan t -i c 1 +csc t , ds (1 — CSC t)(— esc t cot t) — (1 + CSC t)(csc t cot t) ID. S— 1 _ csct =? dt - (1-csct) 2 —esc t cot t + CSC 2 t cot t — CSC t cot t — CSC 2 t cot t —2 CSC t cot t (1-csct) 2 16. s (1 -CSC t) 2 sint . ds _ (1 - cos t)(cos t) - (sin t)(sin t) cos t - cos 2 1 - sin 2 t _ cos t - 1 1 — cos t dt 1 (1 -cos t) 2 (1— cost) 2 (1— cost) 2 1 — COS t COS t — 1 17. r = 4 - 2 sin =>■ % = - (0 2 j- g (sin 0) + (sin 0)(20)) = - (6» 2 cos 6 + 26 sin 9) = -6(6 cos 9 + 2 sin 9) 18. r = 6 sin 6 + cos 6 => f g = (6 cos 9 + (sin 0)(1)) - sin 9 = 9 cos 6 19. r = sec 6 esc 6 => % = (sec 0)(-csc 6 cot 0) + (esc 0)(sec tan 0) = (^i ) ( i ) (coi|) + ( i ) ( i ) (ml) = ^l + l = S ec 2 - esc 2 V cos 8 J \sm 9 / \ sm9 J \ sm 9 / V cos # / V cos 9 / sin 2 # cos 2 9 20. r = (1 + sec 0) sin =^ ^ = (1 + sec 0) cos + (sin 0)(sec tan 0) = (cos + 1) + tan 2 = cos + sec 2 t 21 - p = 5 + ciq = 5 + tan q =► o| = sec2 q 22. p = (1 + esc q) cos q => ^ = (1 + esc q)(— sin q) + (cos q)(— esc q cot q) = (—sin q — 1) — cot q = —sin q — esc q Copffigl (c) 1 Pearson Etation, Inc., publishing as Pearson Addison-Wesle Section 3.4 Derivatives of Trigonometric Functions 143 23. p sin q + cos q cos q dp (cos q)(cos q — sin q) — (sin q + cos q)(— sin q) dq cos z q — cos q sin q + sin^ q + cos q sin q 1 J n ts 2 n ens 2 n VI cos z q cos 2 q r*A tan q _. dp (1 + tan q) (sec 2 q) — (tan q) (sec 2 q) sec 2 q + tan q sec 2 q — tan q sec 2 q sec 2 q P — 1 + tan q dq — (1 + tan q) 2 — (1 +tanq) 2 — (1 +tanq) 2 25. (a) y = esc x => y' = —esc x cot x =4- y" = — ((esc x) (—esc 2 x) + (cot x)(— esc x cot x)) = esc 3 x + esc x cot 2 x = (esc x) (esc 2 x + cot 2 x) = (esc x) (esc 2 x + esc 2 x — 1) = 2 esc 3 x — esc x (b) y = sec x => y' = sec x tan x =>■ y" = (sec x) (sec 2 x) + (tan x)(sec x tan x) = sec 3 x + sec x tan 2 x = (sec x) (sec 2 x + tan 2 x) = (sec x) (sec 2 x + sec 2 x — 1) = 2 sec 3 x — sec x 26. (a) y = —2 sin x =^ y' = —2 cos x =^ y" = — 2(— sin x) = 2 sin x => y'" = 2 cos x =4> y^ = —2 sin x (b) y = 9 cos x => y' = —9 sin x =$■ y" = —9 cos x =$■ y'" = — 9(— sin x) = 9 sin x =$■ y( 4 ' = 9 cos x 27. y = sin x => y' = cos x =4- slope of tangent at x = — 7r is y'(— 7r) = cos (— 7r) = — 1; slope of tangent at x = is y'(0) = cos (0) = 1; and slope of tangent at x = ^ is y' (4^) = cos ^ = 0. The tangent at (— n, 0) is y — = — l(x + tt), or y = —x — 7r; the tangent at (0, 0) is y — = l(x — 0), or y = x; and the tangent at (f,-l)isy = -l. 28. y = tan x =^ y' = sec 2 x =>- slope of tangent at x = — | is sec 2 (— | ) =4; slope of tangent at x = is sec 2 (0) = 1; and slope of tangent at x = | is sec 2 (|) =4. The tangent at (- f , tan(- f )) = (- f , -y^) is y + ^3 = 4(x + f ) the tangent at (0, 0) is y = x; and the tangent at (|, tan (|)) = (f, v / 3)isy-V / 3 = 4(x-f). y=.4x + f-VS y-x y - tan x y-4x-^ + V5 I 29. y = sec x =4- y' = sec x tan x =>• slope of tangent at x = — | is sec (— |) tan (— |) = — 2y 3 ; slope of tangent at x = | is sec (|) tan (|) = \J 2 . The tangent at the point (- § , sec (- f )) = (- f , 2) is y - 2 = -2y/l (x+ |) ; the tangent at the point (f , sec ( ^)) = ( |, \/l\ is y - \fl = V^(x-l). < y = sec x ' \ (-77/3, 2) ! < 2 / ! V <i (tt/4, •&) 1 1 y = ^2x 4 -7T/2 -7T/3 ttIA 77/2 slope of tangent at 30. y = 1 + cos x => y' = —sin x x = — | is —sin ( — f ) = 2 ' s l°P e of tangent at x = y is —sin (y) — 1- The tangent at the point TT 3-\ (-f,l + cos (-!)) = (- f.f) is y — ^3 COS (¥)) ; the tangent at the point ^,1) isy-1 =x- f V3/ it\ '--2-( X + 3) (3k/2,1) Copyright (c) 2006 Pearson Education 144 Chapter 3 Differentiation 3 1 . Yes, y = x + sin x =>• y' = 1 + cos x; horizontal tangent occurs where 1 + cos x = =>• cos x = — 1 => X = 7T 32. No, y = 2x + sin x => y' = 2 + cos x; horizontal tangent occurs where 2 + cos x = are no x-values for which cos x = —2. cos x = —2. But there 33. No, y = x — cot x =>• y' = 1 + esc 2 x; horizontal tangent occurs where 1 + esc 2 x = are no x-values for which esc 2 x = — 1. 1. But there 34. Yes, y = x + 2 cos x =>• y' = 1 — 2 sin x; horizontal tangent occurs where 1 — 2 sin x = =>• 1 = 2 sin x i = sinx =>• x=Jorx=4r J DO 35. We want all points on the curve where the tangent line has slope 2. Thus, y = tan x => y' = sec 2 x so that y' = 2 => sec 2 x = 2 =>■ sec x = ± \/2 =^> x = ± |. Then the tangent line at (?, l) has equation y— 1=2 (x— |); the tangent line at (- f , -l) has equation y + 1 = 2 (x + f ) . 36. We want all points on the curve y = cot x where the tangent line has slope — 1 . Thus y = cot x => y' = —esc 2 x so that y' = — 1 =4> —esc 2 x = — 1 =>■ esc 2 x = 1 => csc x= ±1 =>• x=|. The tangent line at (|, 0) is y = — x + | . 37. y = 4 + cot x - 2 csc x =4> y' = -csc 2 x + 2 csc x cot x = - ( J-) C~ 2cosx ) ^ ^ V sin x / V sin x / (a) When x = |, then y' = — 1; the tangent line is y = — x + | + 2. (b) To find the location of the horizontal tangent set y' = => 1—2 cos x = then y = 4 — y 3 is the horizontal tangent. x = | radians. When x = |, 38. y = 1 + \/2cscx + cotx => y' = - y^csc x cot x - csc 2 x = - (J^) ( y ^f ±1 ) (a) If x = j, then y' = —4; the tangent line is y = — 4x + n + 4. (b) To find the location of the horizontal tangent set y' = '2 cos x + 1 = x = 2s radians. When x = ; ?, then y = 2 is the horizontal tangent. 39. lim sin (i - i) = sin (| - |) = sin = 40. lim ^ y 7 ! + cos(7r csc x) = </l + cos (7r csc (— f)) = yl + cos (w • (—2)) = y 2 41. lim sec [cos x + 7r tan ( 4 s ^ c - ) — 1] = sec [cos + ir tan ( 4 — Q ) — l] = sec [l + w tan (|j — 11= sec 7T = — 1 Copffigl (c) 1 Pearson Ediratio^, k, publishing as Pearson Addison-Wesle Section 3.4 Derivatives of Trigonometric Functions 145 42. lim sin ( t ?r+t ™ x ) = sin ( , 7r +^"° a ) x j, q V tan x— 2 sec x / V tan 0—2 sec / 43. lim tan (l - sal) = tan (l - lim ^ = tan(l - 1) = t-»o v ' ' V t->0 l / 44. lim cos (4^ -> V Sln e ) = COS I 7T lim -/-a ) = COS I 7T • ^TTT ) = cos (w - t) = — 1 45. s = 2 - 2 sin t => v = ^ at -2 cos t =>• a = ^ = 2 sin t =^>j=^ = 2 cos t. Therefore, velocity = v (|) - V 2 m/sec; speed = |v (|) | = y 2 m/sec; acceleration = a (|) = y 2 m/sec 2 ; jerk = j (?) = y2 m/sec 3 . 46. s = sin t + cos t =>• v = ^ = cos t — sin t =>■ a = $ at at da -sin t — COS t =>- J — h velocity = v (? ) =0 m/sec; speed = |v (?) | =0 m/sec; acceleration = a (?) = — y 2 m/sec 2 ; jerk = j (|) = m/sec 3 . cos t + sin t. Therefore 47. lim A f(x) = lim n s -^ = lim n 9 (^) (^) = 9 so that f is continuous at x = =>• lim n f(x) = f(0) x -> x^ => 9 = c. x^O x^O lim_ g(x) = lim_ (x + b) = b and lim g(x) = lim cos x = 1 so that g is continuous at x = =>• lim_ g(x) x^O x^O x^0 + x ^ 0+ " x^O = lim g(x) => b = 1. Now g is not differentiable at x = 0: At x = 0, the left-hand derivative is x — > 0+ ^ (x + b)| = 1, but the right-hand derivative is -^ (cos x)| _ = — sin = 0. The left- and right-hand derivatives can never agree at x = 0, so g is not differentiable at x = for any value of b (including b = 1). g^ggg (cos x) = sin x because f^ (cos x) = cos x => the derivative of cos x any number of times that is a multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 = 249 -4 + 3 d\"" (cos x) £s ;|ss!je (cos x) = £s (cos x) = sin x. 50. (a) y = sec x l COS X dy _ (cosx)(0)-(l)(-sinx) dx (cos x) 2 => 4- (sec x) = sec x tan x (b) y = esc x d l sin x dy _ (sinx)(0)-(l)(cos x) _ -cos x dx (sin x) 2 L = (i)(M) x V cos x / V cos x / / cos X \ V sin x / ' -1 \ Ccosx' v sin x / sec x tan x -esc x cot x => ^ (esc x) = —CSC x cot X (c) y = cot x cos X sin x dy (sin x)(— sin x) — (cos x)(cos x) dx (sin x) 2 — sin x— cos x —1 =>• 4- (cot x) = —CSC 2 X 51. As h takes on the values of 1, 0.5, 0.3 and 0. 1 the corresponding dashed curves of y = Sln (x ± J — !HL5 g et closer and closer to the black curve y = cos x because 4- (sin x) = lim Mn . Sln x = cos x. The same } dx h - h is true as h takes on the values of —1, —0.5, —0.3 and —0.1. Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 146 Chapter 3 Differentiation 52. As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y cos (x + h) — cos x get closer and closer to the black curve y = — sin x because 4- (cos x) = lim co , — ^^ = —sin x. The 3 dx h - h same is true as h takes on the values of — 1, —0.5, —0.3, and —0.1. 53. (a) £ -\ /f- A 5 \ / -7T ij -0.5 \ n 1 2jr -J -1 \^/ The dashed curves of y sin(x + h) — sin(x — h) 2h are closer to the black curve y = cos x than the corresponding dashed curves in Exercise 5 1 illustrating that the centered difference quotient is a better approximation of the derivative of this function. (b) /°=\ X r \ /' ~\\ / o\s / \ -7T -0.5 -1 \ / 27T The dashed curves of y = — > ; 2h -^ ' are closer to the black curve y = — sin x than the corresponding dashed curves in Exercise 52 illustrating that the centered difference quotient is a better approximation of the derivative of this function. 54. lim h^0 |0 + hl-|0-h| lim x^0 - h| lim 2h x _, 2h though the derivative of f(x) = |x| does not exist at x the limits of the centered difference quotient exists even = 0. 55. y = tan x => y' = sec 2 x, so the smallest value y' = sec 2 x takes on is y' = 1 when x = 0; y' has no maximum value since sec 2 x has no largest value on (— |, |) ; y' is never negative since sec 2 x > 1. Copyright (c) 2006 Pearson Education Section 3.4 Derivatives of Trigonometric Functions 147 56. y = cot x =>- y' = —esc 2 x so y' has no smallest value since —esc 2 x has no minimum value on (0, 7r); the largest value of y' is — 1, when x = | ; the slope is never positive since the largest value y' = —esc 2 x takes on is — 1. 57. y - Mn v lim 5»x x^O x at y — appears to cross the y-axis at y = 1, since 1 ; y = ^^ appears to cross the y-axis 2, since lim 5M>2x = 2; y = sin4x appears to cross the y-axis at y = 4, since lim However, none of these graphs actually cross the y-axis since x = is not in the domain of the functions. Also, lim ^^ 5, lim sin(— 3x) -3, and lim ^ x^ ' sin(— 3x) kx = k => the graphs of y = ^5, y = >^-^>, and y = ^^ approach 5, —3, and k, respectively, as x — > 0. However, the graphs do not actually cross the y-axis. y ■ (sin 4x)/x y = (sin 2x)/x y = (sinx)/x 58. (a) h sin h h (^) (^) 1 .017452406 .99994923 0.01 .017453292 1 0.001 .017453292 1 0.0001 .017453292 1 lim lim >( h -ii) lim h->0 '("-Tin) lim ^-, ISO (converting to radians) h cos h— 1 h 1 -0.0001523 0.01 -0.0000015 0.001 -0.0000001 0.0001 lim £2^" h^O h 0, whether h is measured in degrees or radians. (0 = h • Tin) (c) In degrees, -jj (sin x) = lim (sin x • ^ u . n V h lim sin (x + h) — sin x •0 (sinx)(0) + (cos x) (t| + lim (cos x i- (sin x cos h + cos x sin h) — sin x sin h ^ (c\ n v \ U-m /"cosh—i (sin x) • lim twO (cosx)- lim m±\ 180/ COS X (d) In degrees, 4- (cos x) = lim W B dx V > h _> Q " 180 cos (x + h) — cos x h lim (cos x cos h — sin x sin h) — cos x lim (cos x)(cos h — 1) — sin x sin h (cos x) lim h->0 (e) ^(sinx) _d_ dx V 180 h — n - lim (cosx- £2iA^i) - lim (sinx-^) h h^O v h > h->0 v h ; £2^1) - (sin x) h lim o (^) = (cos x)(0) - (sin x) (^) = d_ dx 180 cos x; cosx ) = - (tm) 2 sin x ; a& ( sin x > = ^ (- (ifo) 2 sin x ) = -(ifo) ; £ (cos x) = £ (- jfg sin x) = - (yffj) 2 cos x; £, (cos x) = £ (- (^f cos x) = {^f sin x Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 148 Chapter 3 Differentiation 3.5 THE CHAIN RULE AND PARAMETRIC EQUATIONS 1. f(u) = 6u - 9 => f (u) = 6 => f (g(x)) = 6; g(x) = ^x 4 ^ g'(x) = 2x 3 ; therefore g = f (g(x))g'(x) = 6 • 2x 3 = 12x 3 2. f(u) = 2u 3 => f (u) = 6u 2 => f (g(x)) = 6(8x - l) 2 ; g(x) = 8x - 1 => g'(x) = 8; therefore g = f'(g(x))g'(x) = 6(8x- l) 2 -8 = 48(8x- l) 2 3. f(u) = sin u =>- f'(u) = cos u => f'(g(x)) = cos (3x + 1); g(x) = 3x + 1 =>• g'(x) = 3; therefore g = f'(g(x))g'(x) = (cos (3x + 1))(3) = 3 cos (3x + 1) 4. f(u) = cos u => f'(u) = -sin u => f'(g(x)) = -sin (^) ; g(x) = f => g'(x) = - §; therefore | = f'(g(x))g'(x) -sin (f)-(f) = ^n(f) 5. f(u) = cos u => f'(u) = —sin u =$■ f'(g(x)) = —sin (sin x); g(x) = sin x =>- g'(x) = cos x; therefore % = f'(g(x))g'(x) = -(sin (sin x)) cos x 6. f(u) = sin u =>- f'(u) = cos u => f'(g(x)) = cos (x — cos x); g(x) = x — cos x => g'(x) = 1 + sin x; therefore dy dx ^ - f'(g(x))g'(x) = (cos(x - cos x))(l + sin x) 7. f(u) = tan u => f'(u) = sec 2 u =4> f'(g(x)) = sec 2 (lOx - 5); g(x) = lOx - 5 => g'(x) = 10; therefore g = f'(g(x))g'(x) = (sec 2 (lOx - 5)) (10) = 10 sec 2 (lOx - 5) ]. f(u) = —sec u =4> f '(u) = —sec u tan u =>• f'(g(x)) = —sec (x 2 + 7x) tan (x 2 + 7x) ; g(x) = x 2 + 7x dy dx g'(x) = 2x + 7; therefore g = f'(g(x))g'(x) = -(2x + 7) sec (x 2 + 7x) tan (x 2 + 7x) 9. With u = (2x + 1), y = u 5 : g = g g = 5u 4 • 2 = 10(2x + l) 4 10. With u = (4 - 3x), y = u 9 : g = g g = 9u 8 • (-3) = -27(4 - 3x) 8 11. Withu=(l-f),y = ^: g = ^g = -7u- 8 -(-i) = (l-f)- 12. Withu= f|-l),y = u- 10 : g = £ £ = -lOir 11 - '* v 2 / ' ' dx du dx 13. Withu= (f +x-i),y u 4. dy _ dy du dx du dx 2 ) = 5(| - 1) ^ 4u 3 -(t + l + ^)=4(f+x-i) 3 (| + l + i) 14. Withu=(f + i),y = ^:g = g£=5u 4 .(I-^) = (f + i) 4 (l-^ 15. With u = tan x, y = sec u: gj = gg gj = ( sec u tan u) (sec 2 x) = (sec (tan x) tan (tan x)) sec 2 x 16. Withu = 7 r-i,y = cotu: g = £g = (-esc 2 u) (4,) = - 4, esc 2 (» - 1) 17. With u = sin x, y = u 3 : g = g f = 3u 2 cos x = 3 (sin 2 x) (cos x) 18. With u = cos x, y = 5tT 4 : g = g g = (-20iT 5 ) (-sin x) = 20 (cos~ 5 x) (sin x) Copfiigl (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle Section 3.5 The Chain Rule and Parametric Equations 149 19. p = y/3=l=(?-t)V* =* ^ = I(3-tr 1 /2. d ( 3_ t) = _i (3 _ t) -i/2 2V3-t 20. q = i/2r - r 2 = (2r - r 2 ) 1/2 => f = \ (2r - r 2 ) _1/2 - | (2r - r 2 ) = \ (2r - r 2 ) _1/2 (2 - 2r) ^ 21. s = #■ sin 3t + #- cos 5t => jj? = 4- cos 3t • £ (3t) + #■ (-sin 5t) • £ (5t) = ^ cos 3t - ^ sin 5t 3tt 5tt at in at v 7 5tt v ' at v ' tt tt = - (cos 3t — sin 5t) 22. s = sin(^) +cos(^) '37rt\ ds COS m ■!(¥)- •!» (¥) ■ I (¥) = ¥ cos (f ) - f sin (f ) f (cos 3f£ - sin 2si) cot 9) 2 (esc 9 + cot 9) 2 23. r = (esc + cot 6y l => % = -(esc 9 + cot 0)~ 2 ^ (esc + cot (9) = ™J> «* <> + <*<? o - ^ '-' --^ ^ ' "-' CSC 9 CSC 9 + cot 9 sec 9 tan 9 + sec 2 9 _ sec 9 (tan 9 + sec 9 tan 9) 2 ~~ (sec 9 + tan 9) 2 24. r = -(sec 9 + tan 9)~ x ^ % = (sec + tan 0)-' 2 ^ (sec + tan 0) = se ( c s f c 7 + 6 25. y = x 2 sin 4 x + x cos -2 x =^> g = x 2 £■ (sin 4 x) + sin 4 x • ^ (x 2 ) + x I- (cos -2 x) + cos -2 x • ^ (x) = x 2 (4 sin 3 x -£ (sin x)) + 2x sin 4 x + x (—2 cos -3 x • ^ (cos x)) + cos -2 x = x 2 (4 sin 3 x cos x) + 2x sin 4 x + x( (—2 cos -3 x) (—sin x)) + cos -2 x = 4x 2 sin 3 x cos x + 2x sin 4 x + 2x sin x cos -3 x + cos -2 x 26. y = - sin -5 x — | cos 3 x =>- y' = - £ (sin -5 x) + sin -5 x-r(-)-|i (cos 3 x) - cos 3 x - -jj- (|) = i (-5 sin~ 6 x cos x) + (sin -5 x) (— 4J — | ( (3 cos 2 x) (-sin x)) - (cos 3 x) (|) = — - sin -6 x cos x — 4 sin -5 x + x cos 2 x sin x — \ cos 3 x x x 2 3 27. y= ^(3x-2V + (4- 5 i 2 )- 1 => g = ^(3x-2)e.A ( 3x-2) + (-l)(4-^)- 2 .A( 4 -^) = ^(3x-2) 6 .3 + (-l)(4-^)- 2 (i)=(3x-2) 6 x^ 4- 1 28. y = (5 - 2x) 6 (5-2x)4 - 3 + H= +o 3 dy dx -3(5 - 2x)- 4 (-2) + i (| + 1) J (- |) = 6(5 - 2x)- 4 - (A) (§ 29. y = (4x + 3) 4 (x + l) -3 => g = (4x + 3) 4 (-3)(x + 1)~ 4 - £ (x + 1) + (x + ir 3 (4)(4x + 3) 3 • £ (4x + 3) = (4x + 3) 4 (-3)(x + 1)" 4 (1) + (x + l)- 3 (4)(4x + 3) 3 (4) = -3(4x + 3) 4 (x + l)" 4 + 16(4x + 3) 3 (x + l)" 3 = W? [" 3 ( 4x + 3 ) + 16 < x + !)1 = ( % 3 + ( i)* + 7) 30. y = (2x - 5)- 1 (x 2 - 5x) G => g = (2x - 5r 4 (6) (x 2 - 5x) 5 (2x - 5) + (x 2 - 5x)°(-l)(2x - 5)" 2 (2) 6 (x 2 - 5x 5 _ 2(x 2 -5x) tl 1 (2x - 5)2 31. h(x) = x tan (2y^x) + 7 => h'(x) = x ^ (tan (2X 1 / 2 )) + tan (2X 1 / 2 ) - £ (x) + = x sec 2 (2X 1 / 2 ) • £ (2X 1 / 2 ) + tan (2X 1 / 2 ) = x sec 2 (2^/x) • 4- + tan (2^/x) = ,/x sec 2 (2-^/x) + tan (2,/x) Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 150 Chapter 3 Differentiation 32. k(x) = x 2 sec (I) =* k'(x) = x 2 £ (sec 1) + sec (I) ■ A. ( x 2) = x 2 sec (I) tan (I) . A. (I) + 2x sec (V - ¥) + 2x sec (!) = 2x sec (!) ~ sec (l) tan (I) 33- fW=(if^) ^f'W = 2( T f^)-^ u+cos d_ f sin > W (2 sin g) (cos g + cos 2 9 + sin 2 g) _ (2 sin 9) (cos 9+ 1) _ 2 sin 9 2 sin 9 (1 + cos 9)(cos 9) - (sin g)(-sin I l+cos9 ' (l+cos9) 2 (l+cos9) 3 (l+cos9) 3 (1+c 1 + cosf\~l 34. g(t)=( i ^r => g'ct) = - (4^) _ — (— sin 2 t — cos t — cos 2 t) _ 1 (1 +cost) 2 ~ 1 + cos t 1 + cos t \ ^ d ( 1 + cos t dt V sin P)=- n 2 1 (sin t)(— sin t) — (1 + cos t)(cos t) ( 1 + cos t) 2 (sin t) 2 35. r = sin (<9 2 ) cos (2d) => % = sin (d 2 ) (-sin 2d) j- g (2d) + cos (2d) (cos (<9 2 )) • ± ((9 2 ) = sin (0 2 ) (-sin 20)(2) + (cos 20) (cos (0 2 )) (20) = -2 sin (0 2 ) sin (20) + 2d cos (2(9) cos (0 2 ) 36. r = (sec v^) tan (I) => % = (sec y/o) ( sec 2 i) (- £) + tan (J) (sec yfd tan ^) (^) = - ^ sec v^ sec 2 (\) + -^ tan (i) sec v^ tan y^ = (sec y^ tan ^0 tan (1) sec 2 (1) 275 52_ 37- q - * fe)^ = cc (^) - | (^) = cos (^) - ^*<^ = - (v^T) • ^^ = cos (^) (t^g) = (^) cos (-^) > g = -esc 2 (*1) - I (*«) = (-esc 2 (*!)) (ts-^M-l) 38. q = cot(^ i 39. y = sin 2 (?rt - 2) => f = 2 sin (vrt - 2) • f t sin (vrt - 2) = 2 sin (Trt - 2) • cos (vrt - 2) • | (71I - 2) = 27r sin (irt — 2) cos (7rt — 2) 40. y = sec = sec 2 7rt =^ -| = (2 sec 7rt) - ^ (sec 7rt) = (2 sec 7rt)(sec 7rt tan 7rt) • 4 (7rt) = 27r sec 2 7rt tan 7rt 2t)- 4 =^> I = -4(1 + cos 2tr 5 • I (1 + cos 2t) = -4(1 + cos 2t)~ 5 (-sin 2t) - | (2t) = ^ i)) = -2(l + cot(i))- 3 .(-csc 42. y=(l+ C ot(§)) 9 /• t N -liiL 4 => | = -4(l+cos2tr 5 -^i | = -2(l + cot(i))- 3 .^( 41. y = (1 +cos ;i))- 3 -^(i + cot(|)) 8 sin 2t ~~ (1 +cos2t) 5 {■2) J ' dt I 2 / (l+cot(|)) 3 43. y = sin (cos (2t - 5)) => ^ = cos (cos (2t - 5)) - | cos (2t - 5) = cos (cos (2t - 5)) - (-sin = -2 cos (cos (2t - 5))(sin (2t - 5)) (2t - 5)) - A (2t - 5) ; 5 sin(i)) 44. y = cos i sin (I)) (cos (I dy (5 sin (I))- | (5 sin (|)) = -sin(5sin(i))(5cos(i))-^(i) — I sin (5 s ^ = 3[l + tanM^)] 2 ^[l+tan4(^)]=3[l+tanM^)] 2 [4tan3( 1 L).d tan( ^ )] »' (T2) sec2 (12) - T2] = [1 + ^ (T2)] 2 K (T2) sec2 (&)] 45. y=[l+tan^); (7t)f-2cos(7t)(-sin(7t))(7) = 12[l+tan 4 ( 1 L)] 2 [tan3( 1 L)sec 2 ( I L) I [1 + cos 46. y = \ [1 + cos 2 (7t)f dy dt 6 I tan4 (ll)] • dt L 1 T lim IT2JJ =J [' • 4 (i> 3 (^ -7 [1 + cos 2 (7t)] 2 (cos (7t) sin (7t)) Copffigl (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle Section 3.5 The Chain Rule and Parametric Equations 151 47. y = (1 +cos(t 2 )) 1/2 at t 2-n-i/2 d 2\\-V 2 t \ (1 + cos (t*))- 1 " • i (1 + cos (t 2 )) = I (1 + cos (f))~^ (-sin (t 2 ) • g (t 2 )) i(l + cos(t 2 ))- 1/2 (sin(t 2 ))-2t=-^ii ^/l + cosCt 2 ) 48. y = 4sin(\/l - v m] , £ = 4 cos ( ^/ 1 + ^ ) - £ ( ^1 + 0) = 4 cos ( yj\ + ) • ^-^ • £ (l + 0) 2 cos (v^+\A) cos^i + yi^) v / iT7t-2v / t \A+^t 49. y =(l + I)-=> y ' = 3(l + ir(-i)=-| r (l + ir=>y=(- JO" £(1 + D -(1 + D -£ (JO = ^ (i + i) (i + 1) 50. y=(l- x /x)" 1 =► y ' = -(l-y x )" 2 (-i X - 1 /2) = Kl-^-V 1 ^ =► y« = i [(1 - ^)- 2 (- Ix-3/ 2 ) +x-i/2(_2) (1 - ^)- 3 (- |x-V2)] = i [f x-3/ 2 (i - y*-)- 2 + x-i (i - ,A)- 3 ] = i x -i (i - .A)- 3 [- |x-v2 (1 _ ^) + 1] = i(l-AA)- 3 (-^ + ^l)=i(l-^- 3 (|-^) 51. y = icot(3x- 1) => y' = -icsc 2 (3x- 1)(3) = -|csc 2 (3x- 1) =>■ y" = (-f)(csc(3x- 1)- ^csc(3x- 1)) = -f csc(3x- l)(-csc(3x- l)cot(3x- 1)- £(3x- 1)) = 2 csc 2 (3x - l)cot(3x- 1) 52. y = 9 tan (f) => y' = 9 (sec 2 (f )) (±) = 3 sec 2 (f) => y" = 3 - 2 sec (f ) (sec (f) tan (f )) (|) = 2 sec 2 (f ) tan (|) 53. g(x) = y^ ={► g'(x) = i => g(l) = 1 and g'(l) = \ ; f(u) = u 5 + 1 => f (u) = 5u 4 =* f (g(l)) = f (1) = 5; therefore, (f o g)'(l) = f (g(l)) • g'(l) = 5 • 2 2 54. g(x) = (1 - x)- 1 =* g'(x) = -(1 - x)- 2 (-l) = ^ =* g(-l) = i and g'(-l) = i ; f(u) = 1 - 1 => f'(u)=i => f'(g(-l)) = f (|) =4; therefore, (fog)'(-l) = f'(g(-l))g'(-l) = 4-i = l 55. g(x) = 5^ => g'(x) = jjj => g(l) = 5 and g'(l) = f ; f(u) = cot (f ) => f'(u) = -esc 2 (f ) (§) =2 csc 2 fZLH^ 10 csl " WO/ f'(g(l)) = f'(5) is esc 2 (f ; 10 ; therefore, (f o g)'(l) = f'(g(l))g'(l) = - * 10 2 56. g(x) = 7rx => g'(x) = 7r =^> g (|) = | and g' Q) = ir; f(u) = u + sec 2 u =>• f'(u) =1+2 sec u • sec u tan u = 1+2 sec 2 u tan u => f'(g (1)) = f (|) = 1 + 2 sec 2 | tan f = 5; therefore, (fog)' (1) = f'(g (1)) g' (I) = 5^ 57. g(x) = 10x 2 + x + 1 => g'(x) = 20x + 1 => g(0) = 1 and g'(0) = 1; f(u) 2u u 2 +l f'(u) (u 2 + l)(2)-(2u)(2u) (u2 + l) 2 =f^±| =* f'(g(0)) = f'(l) = 0; therefore, (f o g)'(0) = f'(g(0))g'(0) = 0-1=0 58. g(x)=i-l => g'(x) = -2 => g(-l) = 0andg'(-l) = 2;f(u)=(^j) 2 =* f'(u) = 2 (^j) * (^j) = 2 (£*) ■ (n + T + "i ( r lxl) = 2 i^W = £w =► f '(g(-D) = f '(°) = - 4 = therefore ' (f og)'(-l) = f'(g(-l))g'(-l) = (-4)(2) = -8 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 152 Chapter 3 Differentiation 59. (a) y = 2f(x) =* g = 2f'(x) =* gl 2f'(2) = 2(i) (b) y = f(x) + g(x) => | = f (x) + g'(x) => | = f (3) + g'(3) = 2^ + 5 (c) y = f(x)-g(x) g = f(x)g'(x) + g(x)f (x) dy dx f(3)g'(3) + g(3)f (3) = 3 • 5 + (-4)(2tt) = 15 - 8tt (d) y = fJ r\ v ' J g(x) dy _ g(x)f'(x)-f(x)g'(x) dx [g(x)]2 dy dx| , dy I g(2)f'(2)-f(2)g'(2) _ (2)(l)-(8)(-3) _ 37 [g(2)p 22 6 (e) y = f(g(x)) => g = f (g(x))g'(x) -- dx (f) y = (f(x))V2 ^ I = 1 (f(x))-i/a . f ( X ) f (g(2))g'(2) = f (2)(-3) = i (-3) = -1 f'(x) dy dx f'(2) (i) sfl dx 2^^^ *w 2 yftx> "^ dx l x=2 2 \/f(2J 2x/8 6\/8 12-/2 24 (g) Y = (gWr 2 => g = -2(g(x))- 3 • g'(x) => 1 1 = -2(g(3))- 3 g'(3) = -2(-4)- 3 • 5 _5_ 32 (h) y = ((f(x)) 2 + (g(x)) 2 ) 1/2 g = i ((f(x)) 2 + (g(x)) 2 ) 1/2 (2f(x) - f (x) + 2g(x) • g'(x)) dy dx ->2\-l/2 I x=2 5 3\/l7 1 ((f(2)) 2 + (g(2)) 2 ) ^ (2f(2)f (2) + 2g(2)g'(2)) I Cq2 2 ,2N-l/2 + 2 2 ) i/2 (2-8-i+2-2-(-3)) 60. (a) y = 5f(x)-g(x) g = 5f (x) - g'(x) => g 5f'(l)-g'(l) = 5(-i)-(f) = l (b) y = f(x)(g(x)) 3 => g = f(x) (3(g(x)) 2 g'(x)) + (g(x)) 3 f (x) =* gl = 3f(0)(g(0)) 2 g'(0) + (g(0)) 3 f (0) 3(l)(l) 2 (i)+(l) 3 (5) = 6 (c) y f(x) dy _ (g(x)+l)f'(x)-f(x)g'(x) . dy g(x)+l ^ dx (g(x)+l) 2 ( -4+l)(-l)-(3)(-|) _ (-4+1)2 dx (g(l)+l)f'(l)-f(l)g'(l) (g(l) + l) 2 (d) y = f(g(x)) dy f (g(x))g'(x) ={► g = f'(g(0))g'(0) = f'(l) Q) = (- |) (!)=-£ (e) y = g(f(x)) =* g = g'(f(x))f (x) =► gl = g'(f(0))f (0) = g'(l)(5) 40 3 (f) y=(x n +f(x)) 2 => g = -2(x n +f(x)) 3 (llx 10 + f'(x)) dy dx I) (5) -2(l + f(l))- 3 (ll+f'(l)) -2(i + 3r3(ii-i) = (-| I )(f) (g) y = f(x + g(x)) => g = f (x + g(x)) (1 + g'(x)) => g f'(0 + g(0))(l + g'(0))=f'(l)(l + i) l)(!) 61- £ = &■£: s = cose => § = -sin< -sin(f ) = 1 so that | dfl dt 1-5 = 5 62- g = g-£:y = * = x 2 + 7x-5 => g = 2x + 7 Si =9sothat| = g.|=9-i=3 dx x=l = x, we should get ^ = 1 for both (a) and (b): £ = 5; therefore, g dx ' ' dx 63. Withy ,,.. Jx . , ..... (a) y = f + 7 => g = i ; u = 5x - 35 => | = 5; therefore, g = g-g = j-5 = l,as expected r 1 => g = -<* - i)- 2 (D = o^V : therefore 1 = 1-1 /_ i\2 l 1 ■ * du — 5 ' "" -'-' "^ Cb) Y = i + ^ =* g = -^;u = (x-l)- = -r ■ , ~\v = ^ — -1 ,.2 • , "l,, = (x — l) 2 • -. — ^ = 1, again as expected U 2 (X— l) 2 ((x-1) -1 ) (x-l)2 V ' (X— 1)2 ' ° r With y = x 3 / 2 , we should get g = § x 1 / 2 for both (a) and (b): = U 3^ g =3u 2. u=v /^ | = _^ ;therefore) g = g.du = 3u2 ._^ = 3 ( ^ )2 ._^ = |^ (a) y as expected. dZ = 1 ; U = X 3 du 2^/u again as expected. (b) y = s/u f = 3x 2 ; therefore, g = g ■ £ = i dx dx du dx 2a/ u dy du _ _J_ o 2 _ 1 dx 2^ 2V5 _ 1 o v 2 _ 3 1/2 — n ATT JA ~~ 9 A Copffigl (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 3.5 The Chain Rule and Parametric Equations 153 dx (2 sec 2 f) E af . n 2 7TX 2 scl ' 4 65. y = 2tan(f) = | sec 2 (|) = 7r =^ slope of tangent is 2; thus, y(l) = 2 tan (|) given by y — 2 = 7r(x — 1) =>■ y = 7rx + 2 — 7r (a) I 2 and y'(l) tangent line is (b) y' = | sec 2 (^) and the smallest value the secant function can have in— 2<x<2isl =4> the minimum value of y' is § and that occurs when § = § sec 2 (™) =^ 1 = sec 2 (™) => ± 1 = sec (™) =4> x = 0. 66. (a) y = sin 2x => y' = 2 cos 2x =4> y'(0) = 2 cos (0) = 2 =£- tangent to y = sin 2x at the origin is y = 2x; y = —sin (|) =>■ y' = — \ cos (|) =>■ y'(0) = — \ cos = — \ =^ tangent to y = —sin (|) at the origin is y = — \ x. The tangents are perpendicular to each other at the origin since the product of their slopes is 1. (b) y = sin(mx) =4> y' = mcos(mx) =4> y'(0) = mcosO = m; y = —sin ( — ) => y' = — — cos (-) => y'(0) = — — cos (0) = — — . Since m -(——)= —1, the tangent lines are perpendicular at the origin. (c) y = sin(mx) => y' = mcos(mx). The largest value cos (mx) can attain is 1 at x = =>- the largest value y' can attain is |m| because |y'j = |m cos (mx)| = |m| |cos mx| < |m| • 1 = |m| . Also, y = —sin (-) => y' = — — cos (-) =$■ ly'l = I— cos (— ) I < I — I Icos ( — ) I < A => the largest value y' can attain is I — I . J m Vm/ I-' I Im Vm/I — Imll Vm/I — |m| to J I ml (d) y = sin (mx) =4> y' = m cos (mx) =4> y'(0) = m =$■ slope of curve at the origin is m. Also, sin (mx) completes m periods on [0, 27r]. Therefore the slope of the curve y = sin (mx) at the origin is the same as the number of periods it completes on [0, 2tt]. In particular, for large m, we can think of "compressing" the graph of y = sin x horizontally which gives more periods completed on [0, 2ir], but also increases the slope of the graph at the origin. 67. x = cos 2t, y = sin 2t, < t < 7r => cos 2 2t + sin 2 2t = 1 => x 2 + y 2 = 1 68. x = cos (n — t), y = sin (n — t), < t < tt =^ cos 2 (V - t) + sin 2 (tt - t) = 1 => x 2 + y 2 = l,y > y 2 * 2 + /=l l = irllf \(=0 -2 -\\ -I -2 /' 2 2 2 x + y -1 1 .J- n/2 69. x = 4 cos t, y = 2 sin t, < t < 2n 16 cos 2 1 i 4sin 2 t 16 1 16 70. x = 4 sin t, y = 5 cos t, < t < 2ir 16 sin 2 t i 25 cos 2 t 16 25 1 *- + 2- = 1 16 T 25 1 Copyright (c) 2006 Pearson Education 154 Chapter 3 Differentiation 71. x = 3t, y = 9t 2 , -oo < t < oo => y = x 2 / t y = * 2 / i<o\ ^ 1 f t>0 i 72. x = -Vt,y = t, t > => x or y = x 2 , x < ,t>o x— Vy H i 1 1- f? ■ ■ X 73. x = 2t - 5, y = 4t - 7, -oo < t < oo =>■ x + 5 = 2t =4> 2(x + 5) = 4t =>■ y = 2(x + 5) - 7 =>• y = 2x + 3 74. x = 3 - 3t, y = 2t, < t < 1 =>■ § = t =>• x = 3-3(|) => 2x = 6-3y => y = 2 - | x, < x < 3 . . 4 / / = 5/2 - j = 2x + 3 J =7/4. /-l -1 -2 -3 -4 12 3 4 y-2-(2x/3) 1-1 1 — I — I — I 1 — t-^ — I- 75. x = t, y = \f\ -t 2 , — 1 < t < => y = \f\ - x 2 y = A/l-j: 2 76. x = v/t+ 1, y = yfx, t > =4> y 2 = t => x = vV + 1, y > "-Vy^T . \1 -\ — i — i — t-X y t>o " l ! ■ X 77. x = sec 2 t - 1, y = tan t, - § < t < | =>• sec 2 1 — 1 = tan 2 1 => x = y 2 y / V 3 < < < ir/2 2 1 f = * = / -2 -1 -1 -2 I 1 2 3 4 ' A -3 -f S ,<0 78. x = - sec t, y = tan t, - f < t < | => sec 2 t - tan 2 1 = 1 =4> x 2 - y 2 = 1 v 0<t<*/2 ..2 2 . ,' x -y .1 / Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 3.5 The Chain Rule and Parametric Equations 155 79. (a) x = a cos t, y = — a sin t, < t < 27r (b) x = a cos t, y = a sin t, < t < 2ir (c) x = a cos t, y = — a sin t, < t < 4ir (d) x = a cos t, y = a sin t, < t < 4n 80. (a) x = a sin t, y = b cos t, | < t < ^ (b) x (c) x (d) x b sin t, < t < 2tt a cos t, y = a sin t, y = a cos t, y = b sin t, < t < 4ir b cos t, § < t < 9 f 81. Using (—1, —3) we create the parametric equations x = — 1 + at and y = — 3 + bt, representing a line which goes through (—1, —3) at t = 0. We determine a and b so that the line goes through (4, 1) when t = 1. Since 4=— l+a=^>a=5. Since 1 = — 3 + b =^b = 4. Therefore, one possible parameterization is x = — 1 + 5t, y = -3 - 4t, < t < 1. 82. Using (—1, 3) we create the parametric equations x = — 1 + at and y = 3 + bt, representing a line which goes through (— 1, 3) at t = 0. We determine a and b so that the line goes through (3, —2) when t = 1. Since 3 = — 1 + a =^ a = 4. Since —2 = 3 + b =4> b = —5. Therefore, one possible parameterization is x = — 1 + 4t, y = — 3 — 5t, < t < 1. 83. The lower half of the parabola is given by x = y 2 + 1 for y < 0. Substituting t for y, we obtain one possible parameterization x = t 2 + l,y = t, t < 0. 84. The vertex of the parabola is at (—1, —1), so the left half of the parabola is given by y = x 2 + 2x for x < —1. Substituting t for x, we obtain one possible parametrization: x = t, y = t 2 + 2t, t < —1. 85. For simplicity, we assume that x and y are linear functions of t and that the point(x, y) starts at (2, 3) for t = and passes through (-1, -1) at t = 1. Then x = f(t), where f(0) = 2 and f(l) = -1. Since slope = % = z jE^ = -3, x = f(t) = -3t + 2 = 2 - 3t. Also, y = g(t), where g(0) = 3 and g(l) = -1. Since slope =% = =££ = -4. y = g(t) = -4t + 3 = 3 - 4t. One possible parameterization is: x = 2 — 3t, y = 3 — 4t, t > 0. 86. For simplicity, we assume that x and y are linear functions oft and that the point(x, y) starts at (— 1, 2) for t = and passes through (0, 0) at t = 1. Then x = f(t), where f(0) = -1 and f(l) = 0. Since slope =% = ^j^r = 1 > x = f ( l ) = ll + (- 1 ) = -1 + 1 Also, y = g(t), where g(0) = 2 and g(l) = 0. Since slope = ^ = ^ = -2. y = g(t) = -2t + 2 = 2 - 2t. One possible parameterization is: x = — 1 + t, y = 2 — 2t, t > 0. 87. t = \ =>■ x = 2 cos \ = \fl, y = 2 sin \ v^;| -2 sin t, % ' at =>■ m =_ co t£ = dx w 4 =>■ d 2 y dy'/dt esc 2 1 dx 2 — dx/dt — -2 sin t If => x = cos \ = 1 ; tangent line is y d 2 y I dx 2 2 sin 3 1 dy dx dM dx 2 i,y= v^cosf = -\/3 ; tangent line is y — ( — , ) '2= -1 (x —y/2 y/3 . dx 2 ' dt 2 cos t 2 ) or y = - dy dx dy/dt _ 2 cos t dx/dt — 2 sin t COtt 2y2;^=csc 2 t sint, ^ ' dt V~3 **t => | y3[x-(-i)]ory= V / 3x . d/ dt -y/3 sint V~3 cry dx 2 dx dt x ' dt i dy dx I = l-(x-I)or y = x+i;f 4 L dy/dt dx/dt d 2 y * dx 2 27t dy'/dt dx/dt dy dx 1 ; tangent line is 1 <-3/2 _v d^yl 4 L ^ dx 2 Copyright (c) 2006 Pearson Education 156 Chapter 3 Differentiation 90. t = 3 => x = - v / 3+T = -2, y = y / 3(3) = 3; _ _ 3y/t+T _ dy I _ -3y/3 + l (t+1) -l/2 ) | = | (3t) -l/2 dy (l)w-^ /3t dx | (=3 /3(3) dt 2 V-V "^ dx (_l)( t+ l)-l/2 2; tangent line is y — 3 = — 2[x — (—2)] or y = — 2x — 1; d/ _ V^ [- | (t + l)- 1 / 2 ] + 3 Vm [§ (3t)-V 2 ] 3 . <Py dt 3t 2t,/3t\A+T dx2 v2tV3t ,/t+l d 2 y| dx 2 jh) 3 t73t 91. t= -1 => x = 5,y = 1; y — 1 = 1 • (x — 5) or y = dx _ A* dy _ 4f 3 dt — ^ L ' dt ~~ ^ l 4; d/ 2t dy dy/dt 4t^ t 2 dx ~~ dx/dt — ^47 — d 2 y _ dy'/dt _ 2t dx 2 — dx/dt — 4t ? dx| dfyl dx 2 . (— l) 2 = 1; tangent line is 92. t dy dx 3 (!) 4.i 1 1 1 . dx _ 1 2 ' dt ~~ l cos t, -i = sin t ' dt v/3x -l (1 -cost) 2 3 d 2 y I dx 2 1— (!) " (I) (1 — cos t)(cos t) — (sin t)(sin t) V3 ; tangent line is y — | = y3 ( x ~ f + 2 ) + 2;1 d/ _ dt -4 (1-cost) 2 d^v dx-' dy7dt dx/dt dy dx dy/dt dx/dt 93. t dy dx COS 0,y= 1 sin 9- dx z ' dt sin t dy cos t => dy cot f=0; tangent line is y = 2: d/ d 2 y dx 2 dx CSC 2 t COtt dfyl " dx 2 94. t dy dx X = sec 2 t 2 sec 2 1 tan t sec 2 (-f) -1 = l,y cott => 4/ 1 _ 1 2tant ; tan dy dx . I) 1 . dx l ' dt 2 sec 2 1 tan t, $ = sec 2 t cot I l . 2 ; tangent line is y-(-l)=-i(x-l)ory=-ix-!;f 1 csc 2 f . d^y 2 CSC I =? dx2 2 sec 2 t tan t \ COt 3 t d 2 y I dx 2 95. s = A cos (27tbt) => v = f t = -A sin (27rbt)(27rb) = -27rbA sin (27rbt). If we replace b with 2b to double the frequency, the velocity formula gives v = — 47rbA sin (47rbt) =>• doubling the frequency causes the velocity to double. Also v = -27rbA sin (27rbt) =>- a = ^ = -47r 2 b 2 A cos (27rbt). If we replace b with 2b in the acceleration formula, we get a = — 167r 2 b 2 A cos (47rbt) => doubling the frequency causes the acceleration to quadruple. Finally, a = — 47r 2 b 2 A cos (27rbt) =>• j da dt 87r 3 b 3 A sin (27rbt). If we replace b with 2b in the jerk formula, we get j = 64-7r 3 b 3 A sin (4-7rbt) => doubling the frequency multiplies the jerk by a factor of 8. 96. (a) y = 37 sin [|| (x - 101)] + 25 y 37 cos [fa (x - 101)] ' 2tt n v 365 / 747r COS \& (X • 365 .365 101) The temperature is increasing the fastest when y' is as large as possible. The largest value of cos [M= (x — 101)] is 1 and occurs when ^ (x — 101) = => x = 101 => on day 101 of the year ( ~ April 11), the temperature is increasing the fastest. (b) y'(lOl) 365 cos [|L ( ioi -101)] 365 cos (0) 74^ 365 0.64 °F/day 97. s = (1 + 4t) 1 / 2 => v v = 2(1 + 4t) _1/2 => ds _ 1 dt 2 - dv _ — dt " (1 + 4t)~ 1/2 (4) = 2(1 + 4t)- 1 / 2 => v(6) = -- - \ - 2(1 + 4t) _3/2 (4) = -4(1 + 4t)" 3 / 2 2(1 + 4 • 6)" 1 / 2 = | m/sec; a(6)= -4(1+4-6) -3/2 m/sec Copyright (c) 2006 Pearson Education We need to show a dv dt is constant: a Section 3.5 The Chain Rule and Parametric Equations % = %■ • f and ^ dt as dt as k 2, A dv ds ds ' dt dv ds 2\A ■Ws k 2 which is a constant. 157 99. v proportional to -k- => v = -7- for some constant k V s V s dv ds 2s ; J2 Thnc a — dv — dv . ds — 9X . v inus, a— , _ — • — _ — - v dt ds dt ds k k 2 S 3/2 ' ^ ^ (J 2 \s : (-5) =4> acceleration is a constant times \ so a is inversely proportional to s 2 100. Let f = f(x). Then, a dv dv dx dt dx dt dv dx f « = S (f ) * f « = l ( f ( x )) • f W = f '( x ) f ( x )> as required. 101. T = 27r,/t => 5i = 2tt 1 dL ■f- . Therefore, & = 2£ .& = -£- -iL = ^# = 1 . 27rk A /t /gL du dL du ,/gL ^g 2 y g ^ , as required. 102. No. The chain rule says that when g is differentiable at and f is differentiable at g(0), then f o g is differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at so there is no contradiction. 1 03 . The graph of y = (f o g)(x) has a horizontal tangent at x = 1 provided that (f o g)'( 1 ) = =4> f '(g( 1 ))g'( 1 ) = => either f'(g(l)) = or g'(l) = (or both) =>■ either the graph of f has a horizontal tangent at u = g(l), or the graph of g has a horizontal tangent at x = 1 (or both). 104. (fog)'(— 5)<0 =$■ f'(g(-5)) • g'(-5) < =^ f'(g(-5)) and g'(— 5) are both nonzero and have opposite signs. That is, either [f (g(-5)) > and g'(-5) < 0] or [f (g(-5)) < and g'(-5) > 0] . 105. As h — y 0, the graph of y sin 2(x+h)— sin 2x approaches the graph of y = 2 cos 2x because (sin 2x) = 2 cos 2x. lim h->0 sin 2(x+h)— sin 2x dx O.i i r cos f(x + h) 2 l— cos (x 2 ) , the graph of y = — l - ^ ^- L 106. Ash approaches the graph of y = — 2x sin (x 2 ) because [cos (x 2 )] = — 2x sin (x 2 ). lim h->0 cos [(x + h) 2 ] —cos (x 2 ) h _d_ dx Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 158 Chapter 3 Differentiation 107. f = cos t and f = 2 cos 2t dy dx dy/dt dx/dt 2 cos 2t 2(2cos 2 t-l) . cos t ' theng=0 2 (2 cos 2 1- 1) cos t y? =4> 2 cos 2 1 - 1 = => cos t = ±4= => t = f , ^ , ^ , ^ . In the 1st quadrant: t = | => x = sin | - ., y = sin 2 (|) = 1 =>• I -^-, 1 1 is the point where the tangent line is horizontal. At the origin: x = and y = and =>- sin t = => t = or t = 7r and sin 2t = =^ t = 0, 3tt . the origin. Tangents at origin: & - 3 cos 3t 2x and thus t = and t = 7r give the tangent lines at -2x 108. f = 2 cos 2t and d( dy dy/dt 3 cos 3t 3(cos 2t cos t — sin 2t sin t) dx — dx/dt — 2 cos 2t ~~ 2(2cos 2 t-l) _ 3 [(2 cos 2 t- 1) (cos t)- 2 sin t cost sin t] _ (3 cos t) (2 cos 2 t - 1 - 2 sin 2 1) _ (3 cos t) (4 cos 2 t - 3) ~ 2(2cos 2 t-l) ~~ 2(2cos 2 t-l) ~~ 2(2cos 2 t-l) dy =Q (3 cos t) (4 cos 2 1 - 3) =Q 3 C os t = or 4 cos 2 1 - 3 = 0: 3cost = => t=f , ^fand dx 2(2 cos 2 t— 1) 2 ' 2 ; then 4 cos 2 1 - 3 = =>• cos t = ± >/5 t 7T 57T ITT 117T In the 1 st quadrant: t sin 2(1) x/3 and y = sin 3 (|) = 1 =>• ( ^-, 1 ) is the point where the graph has a horizontal tangent. At the origin: x = and y = =4> sin 2t = and sin 3t = => t = 0, | , n, y and t = 0, the tangent lines at the origin. Tangents at the origin: j- 7T 2tt 4_7r 5n 3 ' 3 ' 7r ' 3 ' 3 2 cos § x, and 5Z 2 ' dx t = and t = 7r give dy 3 cos (3tt) 2 cos (27r) 3 X 2 X 109. From the power rule, with y = x 1 / 4 , we get -=- dy j x 3//4 . From the chain rule, dx *'*<^» 1 1 ' 2 X A I x -3/4_ 2 Vv^ in agreement. 1 10. From the power rule, with y = x 3 / 4 , we get -r- = § x 1//4 . From the chain rule, y 2 V x V x 3 \A _ 3 „-l/4 dy dx ^V^vV* £K^) in agreement. 2,/xVx Xa/X bx 1 ' ( x ' 2^ + 2,/x,/x ^) 3y^ 4,/xT/x 111. (a) dg/dt IP- -ir/2 O— w/2 -1 O- TT -o (b) f t = 1.27324 sin 2t + 0.42444 sin 6t + 0.2546 sin lOt + 0.18186 sin 14t (c) The curve of y = 4j approximates y = -J the best when t is not — w, dt 0, f , nor 7r. df/dt Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 3.5 The Chain Rule and Parametric Equations 159 112. (a) dg/dt -IT -IT/2 w/2 Tf (b) f = 2.5464 cos (2t) + 2.5464 cos (6t) + 2.5465 cos (lOt) + 2.54646 cos (14t) + 2.54646 cos (18t) (c) dh/dt _t 111-116. Example CAS commands: Maple : f := t -> 0.78540 - 0.63662*cos(2*t) - 0.07074*cos(6*t) - 0.02546*cos(10*t) - 0.01299*cos(14*t); g := t -> piecewise( t<-Pi/2, t+Pi, t<0, -t, t<Pi/2, t, Pi-t ); plot( [f(t),g(t)], t=-Pi..Pi ); Df := D(f); Dg := D(g); plot( [Df(t),Dg(t)], t=-Pi..Pi ); Mathematica : (functions, domains, and value for tO may change): To see the relationship between f[t] and f [t] in 111 and h[t] in 112 Clear[t, f] f[t_] = 0.78540 - 0.63662 Cos[2t] - 0.07074 Cos[6t] - 0.02546 Cos[10t] - 0.01299 Cos[14t] f[t] Plot[{f[t],f[t]},{t,-7r,7r}] For the parametric equations in 1 13 - 1 16, do the following. Do NOT use the colon when defining tanline. Clear[x, y, t] tO = p/4; x[t_]:=l-Cos[t] y[t_]:=l+Sin[t] pl=ParametricPlot[{x[t], y[t]},{t, -7T, 7r}] yp[t_]:=y'[t]/x'[t] ypp[t_]:=yp'[t]/x'[t] yp[t0]//N ypp[t0]//N tanline [x_]=y[t0] + yp[t0] (x - x[t0]) p2=Plot[tanline[x], {x, 0, 1}] Show[pl, p2] Copffigl (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesle 160 Chapter 3 Differentiation 3.6 IMPLICIT DIFFERENTIATION 1 v _ x 9/4 =, dy _ 9 5/4 i . y — x ^ dx _ 4 x 2. y = X - 3 / 5 =>. g = _| x -8/5 3. y = ^2x" = (2x)V3 =► g = § (2x)- 2 / 3 ■ 2 = ^ 4. y = ^5x" = (Sx) 1 ^ => g = i (5x )-3/4 .5 = 1^ 5. y = 7 v / x + 6 = 7(x + 6) 1 / 2 => | = | (x + 6)- 1 / 2 - 7 2-v/x + 6 6. y = -2x/x^T = -2(x - l) 1 / 2 => g = -l(x - 1)" 1/2 7. y = (2x + 5)- 1 /' 2 => g = - I (2x + 5)- 3 / 2 - 2 = -(2x + 5)- 3 / 2 y = (1 - 6x) 2 / 3 =>■ g = | (1 - 6x)- 1 / 3 (-6) = -4(1 - 6X)- 1 / 3 9. y = x (x 2 + 1) 1/2 => y' = x • i(x 2 + l)~ 1/2 (2x) + (x 2 + 1) 1/2 • 1 = (x 2 + l)" 1/2 (x 2 + x 2 + 1) = ^=±± 10. y = x (x 2 + 1)- 1/2 => y' = x • (-i)(x 2 + l)- 3/2 (2x) + (x 2 + 1)~ 1/2 ■ 1 = (x 2 + l)- 3/2 (-x 2 + x 2 + 1) -- ■ , . , l ( X 2+l) 3 11 s - , 7 /t2 - t 2 / 7 ^> ds - 2 t -5/7 11.;,— V L — i ^ dt — 7 l 12. r -3 = fl-3/4 =, dr 3 0-7/4 13. y = sin ((2t + 5r 2 / 3 ) => f = cos ((2t + 5r 2 / 3 ) • (- f ) (2t + 5ir 5 / 3 - 2 = - f (2t + 5r 5 / 3 cos ((2t + 5)~ 2 -2/3\ 14. z = cos ((1 - 6t) 2 / 3 ) =>- | = -sin ((1 - 6t) 2 / 3 ) - | (1 - 61)- 1 / 3 (-6) = 4(1 - 61)- 1 / 3 sin ((1 - 6t) : ft) 15. f(x) = J 1 ,1/2 -1/2 (l-x 1 / 2 ) 1 ^ => f'(x)=I(l-x 1 / 2 ) _1// (-ix- 1 / 2 ) '(V 1 -^)^ V x ('-v^) 16. g(x) = 2 (2x-V 2 + 1)~ 1/3 =► g'(x) = - | (2X-V2 + l)~ 4/3 - (-l)x- 3 / 2 = I (2X- 1 / 2 + 1)~ 4/ V 3 / 2 17. h(0) = j/l + cos (26») = (1 + cos 20) 1 / 3 =>• h'(0) = I (1 + cos 26») 2 / 3 • (-sin 20) ■ 2 = - § (sin 25) (1 + cos 26>r 2/3 18. k(0) = (sin (0 + 5)) 5/4 =>• k'(0) = § (sin (5 + 5)) 1/4 • cos(0 + 5) = I cos(0 + 5)(sin(0 + 5)) 1 /4 19. x 2 y + xy 2 = 6: Stepl: (x 2 g+y-2x) + (x-2yg+y 2 -l)=0 Step 2: X 2g + 2x y g = -2xy-y 2 Step 3: g (x 2 + 2xy) = -2xy - y 2 dx Step 4: g = ^£ 20. x 3 + y 3 = 18xy =>- 3x 2 + 3y 2 g = 18y + 18x g =>• (3y 2 - 18x) g = 18y - 3x 2 =► g = Jf=| 21. 2xy + y 2 = x + y: Stepl: (2xg+2y)+2yg = l dy dx Cfiojt (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 3.6 Implicit Differentiation 161 Step 2: 2xg+2yg-g = l-2y Step 3: g(2x + 2y-l) = l-2y Step 4: g dx dy _ dx 2x + 2y — 1 1 - 2y 22. x 3 - xy + y 3 = 1 =* 3x 2 - y - x g + 3y 2 g = => (3y 2 - x) g = y - 3x 2 =^| = ^ 23. x 2 (x-y) 2 Step 1: x 2 [2(x - y) (l - g)] + (x - y) 2 (2x) = 2x - 2y g Step 2: -2x 2 (x - y) g + 2y g = 2x - 2x 2 (x - y) - 2x(x - y) 2 Step 3: g [-2x 2 (x - y) + 2y] = 2x [1 - x(x - y) - (x - y) 2 ] Stpn 4- ^ — 2x [ 1 -"( x -y)-( x -y) 2 ] _ x[l-x(x-y)-(x-y) 2 ] _ x (1 - x 2 + xy - x 2 + 2xy - y 2 ) OLCpt. dx — -2x 2 (x-y) + 2y ~~ v -x2r Y _ v i T 2v_ ¥ 8_ y-x 2 (x-y) x^y — x J + y x — 2x 3 + 3x 2 y — xy 2 x 2 y — x 3 +y 24. (3xy + If = 6y => 2(3xy + 7) - (3x g + 3y) = 6 g => 2(3xy + 7)(3x) g - 6 g = -6y(3xy + 7) g [6x(3xy + 7) - 6] = -6y(3xy + 7) dy dx y(3xy + 7) 3xy 2 + 7y x(3xy + 7) - 1 1 - 3x 2 y - 7x 25. y 2 _ x-1 x+1 26. x 2 = S=Z x + y 9 dy _ (x+l)-(x-l) _ 2 . dy _ 1 *J dx (x+l) 2 (x+1) 2 ~^ dx y(x+l) 2 x 3 + x 2 y = x - y => 3x 2 + 2xy + x 2 y' = 1 - y' =4> (x 2 + 1) y' = 1 - 3x 2 - 2xy =>- y' =■ '■■-"-'■■-■-> x 2 +l 27. x = tany => l = (sec 2 y)g => g = O^ = cos 2 y 28. xy = cot (xy) =>• xg + y = -csc 2 (xy)(xg + yj => xg + xcsc 2 (xy)g = -y csc 2 (xy) - y ■ [x + xcsc 2 (xy)] = — y [csc 2 (xy) + l] dy I dx I dy _ -y [csc 2 (xy) + l] dx x[l+ csc 2 (xy)l 29. x + tan (xy) = =S> 1 + [sec 2 (xy)] (y + x g) = => x sec 2 (xy) g = - 1 - y sec 2 (xy) -l -cos 2 (xy) y —cos 2 (xy) — y x sec 2 (xy) x dy dx - 1 — y sec 2 (xy) x sec 2 (xy) a t sin y — *.y —r it v , -<-' a J ' dx ~ J ~ dx v ' ' dx ~ ■> dx cos y — X 31. ysin( y )=l-xy =» y [cos (l) - (-1) £ - g] + S in (l) - g = -x g ■ -y => dy 1" 1 en- (A 1 Hn f 1 ] 1 xl v=t dy " y -y 2 dx[ y C0 HyJ ' Jm \y) ' X J >^dx _i cos (i) +sin (i) +x ysin (i)-cos(l)+xy 32. y 2 cos(i)=2x + 2y=.y 2 [-sin(i).(-l)i.g]+cos(l).2yg = dy [-inf lN | \^vca-(A ">1 " -- dy 2 2 + 2g => dx^ m ^yj 1 -> C0 ^yJ -J -- dx sin (l) +2ycos (l)_ 2 33. fll/2 i r l/2 _ 1 _v 1 fl-1/2 , 1 .-1/2 dr _ n -^ dr [ 1 1 _ -1 . dr _ 2 \/r _ V^ 2^9 v^ Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 162 Chapter 3 Differentiation 34. r - 2^9 = § 2 / 3 + | 3 / 4 => | - -1/2 = -l/3 + -l/4 ^ dr = 0-1/2 + 0-1/3 + 0-1/4 35. sin(r0) = i => [cos (r0)] (r + %) = =^ § [0 cos (r0)] cos (r0) / -r cos (r0) dr _ -r cos (rfl) dfl ~~ 0cos(r0) 36. cos r + cot 9 = rd =4> (-sin r) %. - esc 2 6 = r+9 % => ^ [-sin r - 0] = r + esc 2 =4> ^ = -- 37. x 2 + y 2 = l => 2x + 2yy' = =* 2yy' = -2x => g = y' = - f ; now to find § , £ (y') = £ (- |) ,// _ y(-l)+xy' -y+x - i -y — - since y d 2 y // — y 2 — x 2 — y 2 — (1— y 2 ) — 1 dx : i = y 38. x 2 / 3 + y 2 / 3 = 1 =>• | x- 1 / 3 + | y- 1 / 3 g = => g [| y" 1 / 3 ] = - f x-V3 => y ' ;l/3. (_ 1 y -2/3) y/ + yl/8 (1 x -2/3j ^ xV3.(-ly- 2 /3) f_ !^ J + yl/3 ( | x" 2 / 3 ) dy _ _ x^ dx y- ,1/3 Differentiating again, y" X 2 / 3 ^73 d 2 ! _ 1 -2/3 v -l/3 i 1 v l/3„-4/3 _ ±i dx 2 3 x -Z/O y -l /a + iyi/^X 3x'l/3 ^ 3yl/3 x 2/3 39. y 2 = x 2 + 2x => 2yy' = 2x + 2 =>• _, dfy _ „ _ y 2 - (x+ l) 2 ^ dx 2 — y — y3 y = 2x + 2 = x+i ± •> 2y y ' J „ _ y-(x+l)y' y ; y-(x+Dp±i 40. y 2 - 2x = 1 - 2y =4> 2y • y' - 2 = -2y' => y'(2y + 2) = 2 => y' = -L- = (y + l)" 1 ; then y" -i (y- 1)3 -(y + i)" 2 • y' -(y+D-^y+i)- 1 => S=y"~ - 41. 2,/y" = x - y => y-!/2 y ' = i _ y ' =j>. y ( y -i/a + l) = l => g = y ^ y-1/2+1 - ^+1 . ""^ differentiate the equation y' (y~ 1/2 + l) = 1 again to find y": y' (- \ y" 3 / 2 y') + (y~ 1/2 + l) y" = (y-l/2 + 1 )y = I [y f y -3/2 ^ g = y, » l^Tj y " 3/2 42. xy + y 2 = 1 => xy' + y + 2yy' = =4> xy' + 2yy' = -y =>• y'(x + 2y) = -y =>• y' - — --- • — (y-l/2+l) 2y3/2(y-l/ 2 +l) 3 2(l + y^) 3 ~y • : L - V (x+2y) ' dx 2 J -(x + 2 y )y' + y(l+2y') _ ~(* + 2y) [u+lyi] + y [' + 2 ((I+lyl)] (x + 2y) 2 (x + 2y) 2 2y(x + 2y) - 2y 2 _ 2y 2 + 2xy _ 2y(x + y) (x + 2y) 3 " (x + 2y) 3 ~~ (x + 2y) 3 (^>[y(x + 2y) + y(x + 2y) - 2y 2 ] (x + 2y) 2 43. x 3 + y 3 = 16 => 3x 2 + 3y 2 y' = =4> 3y 2 y' = -3x 2 => y' = - K ; we differentiate y 2 y' = -x 2 to find y": 9 -2x-2yf-4) 2 -2x-?4 y 2 y" + y' [2y - y'] = -2x =* y 2 y" = -2x - 2y [y'] 2 => y" = J^- = — ^ -2xy 3 - 2x 4 d/V I yS ^ dx 2 -32-32 32 44. xy + y 2 = l => xy' + y + 2yy' = => y'(x + 2y) = -y => y' = ^ => y = (x±Mlzgk^)(i±jyO since y| = - 1 we obtain f\ = LMIizlIM = _ l 45. y 2 + x 2 = y 4 - 2x at (-2, 1) and (-2, -1) =» 2y g + 2x = 4y 3 g - 2 =» 2y g - 4y 4x _ /u,3 dy -2-2x Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 3.6 Implicit Differentiation 163 f, (2y - 4y3) -2-2x dy dx x+1 2y 3 -y dy dx -land^ dx 46. (x 2 + y 2 ) 2 = (x-y) 2 at(L0) and (1,-1) => 2 (x 2 + y 2 ) (2x + 2y |) = 2(x - y) (l - &) dx [2y(x 2 + y 2 )-,-(x-y)] = -2x(x 2 + y 2 ) + (x-y)^ g = ^g^^ ' - Jy dy and dy 47. x 2 + xy - y 2 = 1 =>• 2x + y + xy' - 2yy' = =>■ (x - 2y)y' = -2x - y => y / 2x + y . 2y — x ' (a) the slope of the tangent line m = y' | „ 3 = | =>• the tangent line is y — 3 = | (x — 2) (b) the normal line is y — 3 = — ^ (x — 2) => y = — ^ x + y 48. x 2 + y 2 = 25 =>■ 2x + 2yy' = =>• y' = - (a) the slope of the tangent line m = y' | _ _ 4) y ' the tangent line isy + 4= | (x — 3) 25 (b) the normal line is y + 4 = — ^ (x — 3) =>• y x / y / => 2xy 2 + 2x 2 yy' = => x 2 yy' = -xy 2 (a) the slope of the tangent line m = y'| (1 3) = — | => y = 3x + 6 (b) the normal line is y — 3 = — | (x + 1) => y = 3 =>■ the tangent line is y — 3 = 3(x + 1) 50. y 2 - 2x - 4y - 1 = => 2yy' - 2 - 4y' = => 2(y - 2)y' = 2 => y' = -^ ; (a) the slope of the tangent line m = y'| 2 ,. = — 1 =>• the tangent line is y — 1 = — l(x + 2) (b) the normal line is y — 1 = l(x + 2) =>• y = x + 3 -x- 1 51. 6x 2 + 3xy + 2y 2 + 17y - 6 = => 12x + 3y + 3xy' + 4yy' + 17y' = => y'(3x + 4y + 17) = -12x - 3y / _ -12x-3y . y 3x + 4y+17 ' (a) the slope of the tangent line m = y'| , „ = , ? ~ y 7 = f =>• the tangent line is y — = I (x + 1) (b) the normal line is y — = — ? (x + 1) =>■ y=-^x- ^ 52. x 2 - v^xy + 2y 2 = 5 =* 2x - v^xy' - v^y + 4yy' = => y' (4y - \/3x) = V^y - 2x =* y' - - - - - . V / 3y-2x A —2 I (a) the slope of the tangent line m = y' I / R _\ = v ~ y ^ x (b) the normal line is x = \J 3 =>• the tangent line is y = 2 -2y 53. 2xy + 7r sin y = 27r =>■ 2xy' + 2y + 7r(cos y)y' = => y'(2x + 7r cos y) = — 2y =4> y' — ,, (a) the slope of the tangent line m = y'L „ = , ~ y = — | =>■ the tangent line is I (i,f) 2x + 7r cos y (M) - | (x — 1) =4> y=-fx + 7r (b) the normal line is y — | = - (x — 1) => y = -x — - + f Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 164 Chapter 3 Differentiation 54. x sin 2y = y cos 2x =>• x(cos 2y)2y' + sin 2y = — 2y sin 2x + y' cos 2x =>■ y'(2x cos 2y — cos 2x) = -sin 2y - 2y sin 2x^y'= -^ 2y + 2 y S m 2x J J J cos 2x — 2x cos 2y ' (a) the slope of the tangent line m = y' I ,, ,, = sin ^y + 2y sin 2x I v / r & ^ I (j,f J cos 2x — 2x cos 2y /J, jr\ y-| = 2(x-f) =* y = 2x ^ = 2 =>- the tangent line is (b) the normal line is y Hx 55. y = 2 sin (7rx — y) => y' = 2 [cos (7rx — y)] • (n — y') => y'[l + 2 cos (7rx — y)] = 2ir cos (irx — y) / 2tt cos (ttx — y) . J 1+2 cos (7rx — y) ' (a) the slope of the tangent line m = y' | (1 0) = l + 2 "os I*x - ) = 27r => the tangent line is y - = 2tt(x - 1) => y = 2ttx - 2n (b) the normal line is y — = — ^- (x — 1) =>- y = — j- + y- 56. x 2 cos 2 y — sin y = => x 2 (2 cos y)(— sin y)y' + 2x cos 2 y — y' cos y = =4> y' [— 2x 2 cos y sin y — cos y] = -2x cos 2 y4y'= 92 2xcos2y 4 _ ; J J 2x A cos y sin y 4- cos y (a) the slope of the tangent line m = y' | ~ Nk " s N I (0,7r) 2x 2 cos y sin y + cos y => the tangent line is y = -k (b) the normal line is x = 57. Solving x 2 + xy + y 2 = 7 and y = =^ x 2 = 7 =$■ x= ± \/l => ( — \J 7, j and ( a/7, j are the points where the curve crosses the x-axis. Now x 2 + xy + y 2 = 7 =S> 2x + y + xy' + 2yy' = =4> (x + 2y)y' = — 2x — y y' = - 2^+1 ^ m = _ 2x + y ^ ^ x + 2y x + 2y slope at I — v 7, ) is m = %~ = —2 and the slope at I y 7, ) is -j=- = —2. Since the slope is —2 in each case, the corresponding tangents must be parallel. 58. x 2 +xy + y 2 = 7 => 2x + y + x g + 2y g = =* (x + 2y) g = -2x - y =► g = ^ and | = ^ . (a) Solving ^ = =^ — 2x — y = =4> y = — 2x and substitution into the original equation gives x 2 + x(— 2x) + (— 2x) 2 = 7 => 3x 2 = 7 => x= ±a/I and y = T 2-i/ 1 when the tangents are parallel to the x-axis. ■ ■ , - = 7 =^ — = 1 21 ' \ 21 ' =? 4 ' (b) Solving f^=0 =^ x + 2y = =4> y = - § and substitution gives x 2 + x (- §) + ' x ^" - n =>• x = ± 2 J | and y = T yf when the tangents are parallel to the y-axis. 59. y 4 = y 2 - x 2 =^> 4y 3 y' = 2yy' - 2x => 2 (2y 3 - y) y' = -2x => y' y-2y 3 the slope of the tangent line at (4.$)* ^ y-2y3 \(££\ ~ j/fTibZI - i_3 - 2 _ 3 1 I 4 i 1 I 2 S "* 4 1 ; the slope of the tangent line at (4,5) £. is 2\/3 V^ 60. y 2 (2-x) = x 3 => 2yy'(2-x) + y 2 (-l) = 3x 2 => y' - f±H- • m y 2 -3x 2 I 2y(2-x) y - 1 = - \ (X - 1) =4> y = -i x -, ., , s , . , the slope of the tangent line is 2 => the tangent line is y — 1 = 2(x — 1) =4> y = 2x— 1; the normal line is 1 „ 1 3 61. y 4 - 4y 2 = x 4 - 9x 2 => 4y 3 y' - 8yy' = 4x 3 - 18x =>• y' (4y 3 - 8y) = 4x 3 - 18x =4> y / _ 4x 3 - 18x _ 2x 3 - 9x 4y 3 - 8y 2y 3 - 4y Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 3.6 Implicit Differentiation 165 x (2x 2 - 9) y(2y 2 -4) m; (—3,2): m (-3)(18-9) 2(8-4) -f ;(-3,-2): m: 27 . (3,2): m: 27 . (3,-2): m=- 27 62. x 3 + y 3 - 9xy = =4> 3x 2 + 3y 2 y' - 9xy' - 9y = => y' (3y 2 - 9x) = 9y - 3x 2 =>• y' ( a ) y'1,4.2, = I and y' 1 ,2,4) = ?; 9y - 3x 2 3y 2 - 9x 3y-x 2 y 2 — 3x (b) y' = => pEg = => 3y - x 2 = => y = f => x 3 + (f ) - 9x (f ) = => x 6 - 54x 3 = =*• x 3 (x 3 - 54) = =4> x = or x = V54 = 3 3 y / 2 => there is a horizontal tangent at x = 3 V 2 . To find the corresponding y-value, we will use part (c). (c) y 2 — 3x 3y - x 2 => y 2 -3x = => y = ± a/3x ; y = v^x =>• x 3 9xV3x = :? dy => x 3 - 6a/3 x 3 / 2 = => x 3 / 2 (x 3 / 2 - 6\/3) = => x 3 / 2 = or x 3 / 2 = 6^ => x = or x = VT08 = 3 \A • Since the equation x 3 + y 3 — 9xy = is symmetric in x and y, the graph is symmetric about the line y = x. That is, if (a, b) is a point on the folium, then so is (b, a). Moreover, if y' | . b = m, then j/| . a) = ^ . Thus, if the folium has a horizontal tangent at (a, b), it has a vertical tangent at (b, a) so one might expect that with a horizontal tangent at x = y 54 and a vertical tangent at x = 3 y 4, the points of tangency are I v 54, 3 v 4 ) and I 3 y 4, y 54 ) , respectively. One can check that these points do satisfy the equation x 3 + y 3 - 9xy = 0. 63. x 2 - 2tx + 2t 2 2y 3 - 3t 2 2x .,2 dy dx 2x-2t 6y 2 f t ~ 6t = dx dt + 4t = => (2x- 2t)| = = 2x -4t => dx dt " 2x- " 2x- -4t -2t _ x-2t — x-t ; dy dt 6t ~~ 6y 2 " _t r - ; *us g dy/dt ~~ dx/dt _ (?) . " y 2 (> -t) ,-2t) ;t = 2 . + 4 = =^ (x - 2) 2 = => X = 2;t = 2 => 2y 3 - 3(2) 2 = 4 2y 3 = 16 y = 2; therefore -%■ J ' dx 64. x = ^5 - v/i => I = I (5 - Vt) =► (t-i)i -1/2 |t- i/2 ) 2(2- -2) (2) 2 (2- -2(2)) 1 tv/t^5-,/t ;y(t-l)=v/t => y +(t_ 1)| = 1^/8 2\A dy _ yT y _ l-2y\A . , dy dt ~ (t-1) - 2ty/l-2y/i ' dx l-2s\._ 2ly/t — 2\/t -1 4\A l/5 " \A l-2yy/t 2vA(t-l) 4Vt\/5-Vt 2(1-2x^)^5-^ therefore, l-t dy 5- y r A=y r l; t = 4 => y(3) = ^ = 2 i(l-2(2) V / 4)v / 5-V / 4 1-4 14 3 65. x + 2x 3 / 2 = t 2 + t dx dt + 3xV2 I = 2t + l =* (! + 3xl/2 ) I = 2t + ! dx _ 2t+l dT- TT 3^;y^+ T + 2t ^y = 4 V / FTT + y(i)(t+l)-V2 + 2yjF+2t(iy-V2)| = 0^ | yTTT+ ^ + 2^+ (-^) £ = dy _ (yJTT- 2 A> -yyy-4yy^TT t+1 v/y/ dt 2Vt+ fc- 2 v^ thus -y^/y-*y\/tTJ_ dy _ dy/dt _ V2^(i+l) + 2t s /FTT dx dx/dt / 2t + 1 \ ^1+3x1/2; dt " (7F+T+ -Sj) " 2 v /^(t+l) + 2t^Tl ; t = => x + 2x 3 / 2 = =^ x (l + 2x J / 2 ) =0 => x = 0;t = yy / Q+ 1 + 2(0)^ = 4 => y = 4; therefore -4^-4(4)^0 + 1 2^(0+11 + 2(0)^0+7 2(0)+ 1 ~\ l+3(0)V2i Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 166 Chapter 3 Differentiation 66. x sin t + 2x = t t sin t — 2t = y dx sin t + x cos t 9 dx z dt 1 sin t + t cos t (sint + 2)f = — • thus — — sin [ + ' cos ' ~ 2 • 1 — X cos t dx _ dt sin t+2 dt £ ; therefore ■/■ 2 ' dx sin 7T + 7T COS 7T — 2 (h)costt dx -4--S 2 + 7T 1 — X COS t t = 7T =>- X Sin 7T + 2X = 7T sin t + 2 -4 67. (a) if f(x) = | x 2 / 3 - 3, then f'(x) = x" 1 / 3 and f"(x) (b) if f(x) = ^ x 5 / 3 - 7, then f'(x) = | x 2 / 3 and f"(x) = x" 1 / 3 is true i x 4 / 3 so the claim f"(x) = x 1|/3 is false (c) f"(x) = \-^ 3 => f'"(x) = - i x~ 4 / 3 is true (d) if f'(x) = | x 2 / 3 + 6, then f"(x) = x" 1 / 3 is true 68. 2x 2 + 3y 2 also, y 2 = > 4x + 6yy' = 2yy' = 3x 2 => \ 3x 2 2y 2x 3y y' (1,1) = - 2x 3y| 3x 2 3 2y CM) 2 (1,1) and y' | andy'| (U _ n 2x 3y (i-i) 3x^1 2y 2 . — 3 ' l-l) Therefore i (i,-D the tangents to the curves are perpendicular at (1, 1) and (1,-1) (i.e., the curves are orthogonal at these two points of intersection). 69. x 2 + 2xy - 3y 2 = tangent line m = y" y'(2x - 6y) = -2x - 2y x + y 2x + 2xy' + 2y - 6yy' = - , ._.. ,., , _. _, - -, - 3y _ x - — - ' — 1 => the equation of the normal line at (1,1) is y — 1 = — l(x — 1) the slope of the CM) 3y-*lc M) => y = —x + 2. To find where the normal line intersects the curve we substitute into its equation: x 2 + 2x(2 - x) - 3(2 - x) 2 = => x 2 + 4x - 2x 2 - 3 (4 - 4x + x 2 ) = => -4x 2 + 16x - 12 = =^> x 2 — 4x + 3 = => (x — 3)(x — 1) = => x = 3 and y = — x + 2 = — 1. Therefore, the normal to the curve at (1, 1) intersects the curve at the point (3, —1). Note that it also intersects the curve at (1, 1). 70. xy + 2x-y = =>• x^| + y + 2-^ = =>■ % = f±f ; the slope of the line 2x + y = is -2. In order to be parallel, the normal lines must also have slope of —2. Since a normal is perpendicular to a tangent, the slope of the tangent is i . Therefore, y^ = | =>• 2y + 4 = 1 — x =>• x = — 3 — 2y. Substituting in the original equation, y(-3 - 2y) + 2(-3 - 2y) - y = => y 2 + 4y + 3 = => y = -3ory = -l. If y = -3, then x = 3 and y + 3 = -2(x-3) =4> y = -2x + 3. If y = -1, then x = -1 and y + 1 = -2(x + 1) =4> y = -2x - 3. yi-o 71. y=x =>• gj = ;r ■ If a normal is drawn from (a, 0) to (xi,yi) on the curve its slope satisfies ^— - = — 2yi => yi = — 2yi(xi — a) or a = Xj + \ . Since xi > on the curve, we must have that a > \ . By symmetry, the two points on the parabola are (xj , a/x7) and (xi , — a/x+) ■ For the normal to be perpendicular, 'i _a / V a_x i/ -1 => Therefore, ? , ± j) and a (a- Xl )2 _ 3 ~ 4 • 1 =>- Xi = (a — Xi) 2 => Xi *1 i _ Xl ) => X! = \ andyi 72. Ex. 6b.) y = x 1//2 has no derivative at x = because the slope of the graph becomes vertical at x = 0. Ex. 7a.) y = (1 — x 2 ) has a derivative only on (— 1, 1) because the function is defined only on [— 1, 1] and the slope of the tangent becomes vertical at both x = — 1 and x = 1 . 73. xy 3 + x 2 y = 6 _ y 3 + : ~~ 3xy 2 + x 2 *(3y 2 £) .,3 jl v2 dy + 2xy y> -j- x 1 (3xy 2 2xy dx '—^- . a i SO; xy 3 + x 2 y = 6 ^ x ( 3y 2) + y 3 I + x 2 y (2x I) = => I (y 3 + 2xy) — y 3 — 2xy 3xy 2 + x 2 -3xy 2 - dx dy 3xy 2 + x 2 y 3 + 2xy thus p- appears to equal -^ . The two different treatments view the graphs as functions dy Copyright (c) 2006 Pearson Education Section 3.6 Implicit Differentiation 167 symmetric across the line y = x, so their slopes are reciprocals of one another at the corresponding points (a, b) and (b, a). 74. x 3 + y 2 = sin 2 y => 3x 2 + 2y ^ = (2 sin y)(cos y) gf => ^ (2y - 2 sin y cos y) = -3x 2 => 3x 2 —w- ; also, x 3 + y 2 = sin 2 y 3x 2 jjs + 2y = 2 sin y cos y ^ — ■ ?r~ , (ALJKJ. j\. ~r y — axil v — f ~»./v ~r~ ~t~ ^1 — ^- am v ws V — f j~~ — t — o ■> mun j 2 sin y cos y — 2y ' * J J dy J J J dy 5x z ' dy appears to equal A . The two different treatments view the graphs as functions symmetric across the line d7 y = x so their slopes are reciprocals of one another at the corresponding points (a, b) and (b, a). (b) dy _ -3x 2 dx 2y — 2 sin y cos y ; thus 75. x 4 + 4y 2 = 1: (a) y 2 _ 1-x 1 — 4 ±W\-x i dx ±i(i-xT 1/2 (-4x 3 ) ±X' J (l-x^) 1 '' 2 ' differentiating implicitly, we find, 4x 3 + 8y -i = dy _ -4x 3 -4x 3 _ ±x 3 =? * dx ~~ 8y 8(±i/r^ f ) (l-x^) 1 / 2 y' 4 2 V I ) K / ( -l f- -2 -4 ^ y x 4 + 4y*- - 1 76. (x - 2) 2 + y 2 = 4: (a) y = ± v/4 - (x - 2) 2 => g = ± I (4 - (x - 2) 2 r 1/2 (-2(x - 2)) = [4 _^ x _~ 2 ^i/ 2 ; differentiating implicitly, -2(x-2) 2(x-2) + 2y|=0^ g -(x-2) -(x-2) 2y ±(x-2) ± [4 - (x - 2) 2 ] 1/2 [4 - (x - 2) 2 ] 1/2 (b) (x-2) 2 +y 2 = 4 77-84. Example CAS commands: Maple : ql := x A 3-x*y+y A 3 = 7; pt:=[x=2,y=l]; pi := implicitplot( ql, x=-3..3, y=-3..3 ): pi; Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 168 Chapter 3 Differentiation eval( ql, pt ); q2 := implicitdiff( ql, y, x ); m := eval( q2, pt ); tan_line := y = 1 + m*(x-2); p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ): p3 := pointplot( eval([x,y],pt), color=blue ): display( [pl,p2,p3], ="Section 3.6 #77(c)" ); Mathematica : (functions and xO may vary): Note use of double equal sign (logic statement) in definition of eqn and tanline. «Graphics" ImplicifPlof Clear[x, y] {xO, y0}={ l,7i74}; eqn=x + Tan[y/x]==2; ImplicitPlot[eqn,{ x, xO - 3, xO + 3),{y, yO - 3, y0+ 3j] eqn/.{x — » xO, y — » yO} eqn/.{ y -> y[x]} D[%, x] Solve[%, y'[x]] slope=y'[x]/.First[%] m=slope/.{x — * xO, y[x] — > yO) tanline=y==yO + m (x — xO) ImplicitPlot[{ eqn, tanline}, {x, xO - 3, xO + 3},{y, yO - 3, yO + 3}] 3.7 RELATED RATES 1. A = 7rr 2 =>• ^ = 27rr| 2. S = 47rr 2 dS q„~ dr 3. (a) V = 7rr 2 h (c) V = 7rr 2 h dy _ _ r 2 dh dt ~~ ;11 dt dV „J1 dh i o^^U dr w = tit s + 27rrh a (b) V = 7rr 2 h f = 27rrh£ 4. (a) V = ±7rr 2 h (c) _ = 1—2 dh dt 3 dt dY _ l_ r 2 dh dt ~~ t" 1 * dV 3' vrrhl dt (b) V = i7rr 2 h dV 2_.v, dr ^F = S 7 ™ dt 5. (a) ^ = 1 volt/sec (0 f =R(f)+l(f . .. jn l r . ii / i dR _ 1 /dV (b) gj = — 5 amp/sec v ' dt 1 /dV _ tj dl\ . dR _ 1 /dV _ V dl\ - u, - ui u, I I dt R dJ =? dt I V dt I dJ ( d ) W = 2 t 1 _ f (~ I)] = (I) ( 3 ) = I ° hms / sec > R is increasing 6. (a) P = RI 2 => f =I 2 f +2RI^ (b) P = RI 2 =* = f = I 2 § -r 7. (a) s = ^?T7 = (x 2 + y 2 ) V2 (b) s = ^Tf = (x 2 + y 2 ) 1/2 9 RT dI . dR _ _ 2RI dl _ _ 2jf dt "T" ZK ' i dt ^ dt — I 2 dt ~ (c) s = a/x 2 + y 2 => s 2 = x 2 + y 2 8. (a) s = yjx 2 + y 2 + z 2 =4> s 2 = x 2 + y 2 2M dl _ _ 2(f) dl _ 2P dl I 2 dt I 2 dt I 3 dt ds x dx dt ~~ ,/x 2 + y 2 dt ds x dx i y dy h ~ V x2 + y 2 dt \/x 2 + y 2 dt 2s | =2 x| + 2y| =► 2s-0 = 2x| dt 2y| + z 2 =* 2s|=2x£ + 2yf - -2z — zz dt dx dt y dy x dt Cfiojt (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 3.7 Related Rates 169 ds dt x dx \/x 2 + y 2 + z 2 dt (b) From part (a) with 2* = (c) From part (a) with % = y dy , z dz 7x2+^2+12 dt "•" ^/ x 2 + y2 + z 2 dt _s. ds _ _y dy i dt = 2xf y X 2 + y2 + z 2 dt T v / x 2 + y2 + z 2 dt 2y dy dt 2z — ""■ dt dx , y dy , z dz dt ' x dt ' x dt 9. (a) A = i ab sin 9 (c) A = \ ab sin 9 ^ = 1 ab cos 9 f ^ = Iabcos0f (b) A = i ab sin 9 | b sin i da dt db ^ = 1 ab cos 9 § + 1 b sin (9 f dt 2 dt 2 dt 10. Given A = tit 2 , g 0.01 cm/sec, and r = 50 cm. Since ^ = 27rr || , then dAI dt I = 7r cm 7min. 1. Given % dt = — 2 cm/sec dw ' dt = 2 cm/sec, 1 = (a) A = Av => ^ = /? dw * dt + w£ =^ dA dt (b) (c) P = D = 2£ + 2w => dP dt (w 2 2^ + 2 dw dt Vw 2 + P- = dU dt 2*(50)Ug) 12(2) + 5(— 2) = 14 cm 2 /sec, increasing 2(— 2) + 2(2) = cm/sec, constant i(w 2 + f 2 r 1/2 (2w^ + 2/f) dD dt . 5X21+02X^2) _ _ M cm/seC; decreasing v / 25+~T44 Vw 2 + < 12. (a) V = xyz => f = yz | + xz | *y| dS dt dt I (4,3,2) (2y + 2z) | + (2x + 2z) % + (2x + 2y) d| dt _ J" dt (b) S = 2xy + 2xz + 2yz =* f 1 ,4,3,2, = (^X 1 ) + ( 12 )(- 2 ) + (14)(1) = m 2 /sec (c) I = v/x 2 + y 2 + z 2 = (x 2 + y 2 + z 2 ) 1/2 - «" - (3)(2)(1) + (4)(2)(-2) + (4)(3)(1) = 2 m 3 /sec dy i /"1„ i 0,A dz dx dy dj| dt I (4,3,2) '29 dt y/x 2 + y 2 + z 2 dt ^/x 2 + y 2 + z 2 dt W+(^)(- 2 ) + fe)( 1 ) = 0m/SeC dz v / x 2 + y 2 + z 2 dt 13. Given: $ = 5 ft/sec, the ladder is 13 ft long, and x = 12, y = 5 at the instant of time (a) Since x 2 + y 2 169 it dt x dx y dt (5) 12 ft/sec, the ladder is sliding down the wall |xy dA dt (b) The area of the triangle formed by the ladder and walls is A is changing at \ [12( 12) + 5(5)] = - x -f = -59.5 ft 2 /sec. (c) cos0=^ => -sinflf = i-| =* f = _ n ^.^ = _(i) ( 5) = -irad/ S ec D(*f +yf)-Thear ea 14. s 2 = y 2 + X 2 = -614 knots 2s 2x dx 2y£ d§ = I ( x dx , dy\ dt s \^ A dt x J dt ) ds dt 7^ [5(-442) + 12(-481)] 15. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the ,2 _ n.r\(\\2 i „2 _^ ds _ x dx _ 400(25) 500 girl and kite (300) 2 ds dt x dx s dt 20 ft/sec. 16. When the diameter is 3.8 in., the radius is 1.9 in. and 3000 in/min. Also V = 67rr dV dt 12tt(1.9) i 3000 1 0.00767T. The volume is changing at about 0.0239 in 3 /min. dV dt 127T1- 17. V = \ ?rr 2 h, h (a) (b) r dh I dt I h=4 4h 3 = I (2r) = w) do) dr _ 4 dh dt — 3 dt ,2, h(frh 167rh 3 27 90 256ir 4 / 90 3 V 2567T r= f => V 0.1119 m/sec= 11.19 cm/sec 0.1492 m/sec = 14.92 cm/sec dV _ 167rh 2 dh dt 9 dt 15 32tt Copyright (c) 2006 Pearson Education 170 Chapter 3 Differentiation 18. (a) V= |?rr 2 handr 15h V 1 _ M5hW 3^1 — ) h 75;rh 3 4 (b) r -0.0113 m/min = —1.13 cm/min 15h . dr _ 15 dh . dr I _ / 15 \ / -8 \ _ -4 2 dt dt I h=5 2 ) \215t\) 15-tt dV _ 2257rh 2 dh dt 4 dt 0.0849 m/sec dh I 4(-50) -8 dt I h=5 225tt(5) 2 225tt -8.49 cm/sec , -l 19. (a) V=fy 2 (3R-y) => f = f [2y(3R - y) + y 2 (-l)] f =► f = [f (6Ry - 3y 2 )p f => at R = 13 and v " w "■" l " ,v " d ' " -jjj(-6) = £■= m/min (b) The hemisphere is on the circle r 2 + (13 — y) 2 = 169 (c) r = (26y - y 2 ) 1/2 => $ = § (26y - y 2 r 1/2 (26 - 2y) £ = f _ -l 144jt v "'' ~~ 24?r .2_,_na_,A2 _ iAQ -v r= v /26y - y 2 m 13-8 v/26y - y 2 dt dt I y=s ^26-8 - 64 V 24tt / gi m/min 20. If V = | tit 3 , S = 4?rr 2 , and <^ = kS = 4k?rr 2 , then ^ = 47rr 2 | => 4k7rr 2 = 4m 2 | => | = k, a constant. Therefore, the radius is increasing at a constant rate. 21. IfV= | Trr 3 , r = 5, and 100tt ft 3 /min, then ^ = 47rr 2 | dr dt 1 ft/min. Then S = 47rr 2 dS 87rr $ = 87t(5)(1) = 407r ft 2 /min, the rate at which the surface area is increasing. 22. Let s represent the length of the rope and x the horizontal distance of the boat from the dock. (a) We have s 2 = x 2 + 36 = — I = IP (_2) dt I s=io Vl0 2 -36 V ' (b) cos 8 = f =4> - sin 8 % (-2) dt io 2 (^) dx s ds s ds dt x dt \/s 2 - 36 dt ' -2.5 ft/sec. 6 di , r 2 dt ^ dfl dt — 6 r 2 sin 8 - J; rad/sec ■i . Therefore, the boat is approaching the dock at . Thus, r = 10, x = 8, and sin 8 = -^ 23. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal distance between the balloon and the bicycle. The relationship between the variables is s 2 = h 2 + x 2 ds dt 1 A, dh dx'i h f t + x dt ds dt 85 ,\ [68(1) + 51(17)] = lift/sec. 24. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 3, the volume of the coffee is V = 97rh => 2V = 9n dh ^ the rate th coffee is risin ■ dh = l dv = io in/min dt dt ° dt 9tt dt 97T (b) Let h be the height of the coffee in the pot. From the figure, the radius of the filter r : jrh 3 the volume of the filter. The rate the coffee is falling is dh _ 4 dV dt 7rh 2 dt 4 25:, => V = 1 7rr 2 h (-10) = -± in/min. 25. y = QD dy D -1 dQ dt q D -2 dD = i_ (Q) _ M3_ ( _ 2) = 4^_ L/min ^ increasing about 0.2772 L/min 26. (a) dc dt (3x 2 - 12x + 15) § = (3(2) 2 - 12(2) + 15) (0.1) = 0.3. dr dx 9(0.1) = 0.9, ^ 0.9 - 0.3 = 0.6 (b) | = (3x 2 - 12x - 45x- 2 ) | = (3(1.5) 2 - 12(1.5) - 45(1.5r 2 ) (0.05) = -1.5625, | = 70 f = 70(0.05) = 3.5, 3.5 -(-1.5625) = 5.0625 dp dt 27. Let P(x, y) represent a point on the curve y = x 2 and 8 the angle of inclination of a line containing P and the origin. Consequently, tan 8 tan i — = x sec 2 a dfl _ dx dt ~~ dt cos 2 8 f . Since f = 10 m/sec and cos 8\ .=3 y2+x 2 3 2 9 2 +3 2 ■k , we have 1 rad/sec. 28. y = (-x) 1 / 2 and tan 8 tan 6 (-x) 1 - sec 2 f? x 2 dt Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 3 .7 Related Rates 1 7 1 dfl dt EB_ m (-8) (cos 2 (9) (f ) .Now, tan i rad/sec. cos I 2 v/5 COS 2 fl = f Then 29. The distance from the origin is s = y/x ds I dt I (5,12) (x 2 + y 2 )- 1/2 (2xf +2y£) (5.12) y 2 and we wish to find = < 5 X-l) + (12)(-5) = _ 5 ^ V25 + 144 30. When s represents the length of the shadow and x the distance of the man from the streetlight, then s dl dt ds I dx dt ' dt (a) If I represents the distance of the tip of the shadow from the streetlight, then I = s + x (which is velocity not speed) => |atl = ll^t+^tl = llll^l = 5l _5 l = 8 ft/ sec > tne speed the tip of the shadow is moving along the ground. (b) 37 = I ^ = | ( — 5) = —3 ft/sec, so the length of the shadow is decreasing at a rate of 3 ft/sec. 31. Let s = 16t 2 represent the distance the ball has fallen, h the distance between the ball and the ground, and I the distance between the shadow and the point directly beneath the ball. Accordingly, s + h = 50 and since the triangle LOQ and triangle PRQ are similar we have I 30h . " 50 -h =? * 1500 _ on 16t 2 JU 50 - 16t 2 and I 30(50- 16t 2 ) 50-(50-16t 2 ) dl dt 1500 8t? dl I dt I 1500 ft/sec. Ball at time t = sec later SP J a A0 _ 1 ds sec o d( — 132 d( 32. Let s = distance of car from foot of perpendicular in the textbook diagram =4> tan 9 — jf* -264 and 9 = =>• % = —2 rad/sec. A half second later the car has traveled 132 ft £ .2 cos 2 8 ds . ds 132 dt ' dt right of the perpendicular =>■ \8\ = |, cos 9 i and f 2 dt 264 (since s increases) 1(264) 1 rad/sec. 33. The volume of the ice is V 7T4 d dV dt 471T .2 dr — I dt I r=6 thickness of the ice is decreasing at =f- in/min. The surface area is S = 47rr 2 wJ- in./min when ^ 12tt dt dS o^-„ dr * dS I -10in 3 /min, the 10 — y in/min, the outer surface area of the ice is decreasing at y in J /min 34. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between the car and plane => 9 + s 2 = r 2 => % = -^= % =^ %\ = -2- (-160) = -200 mph r dt ,y r 2_9 dt dt I r=5 ,^16 v ' r => speed of plane + speed of car = 200 mph =>■ the speed of the car is 80 mph. 35. When x represents the length of the shadow, then tan 9 = — =>• sec 2 9 ^ We are given that 0.27° I dx I I dt I -x 2 sec 2 dfl dt 80 2^g rad/min. At x = 3| ft/min » 80 . x = 60, cos 9 = 0.589 ft/min 80 dx x 2 dt 7.1 in./min. dx dt at A sec 2 9 M 36. Let A represent the side opposite 9 and B represent the side adjacent 9. tan 9 = g —r ow w d ,, • - - * -10mandB = 20mwehavecosc9 =I ^ = -2=andf =[(i)(-2)-(i|(l))](| -^/sec w -67sec l dA B dt A_ dB B 2 dt '^1 1_\ /4^ v 10 40 J V5/ — -Tfj rad/sec 37. Let x represent distance of the player from second base and s the distance to third base. Then dx dt 16 ft/sec (a) s 2 8100 2s - -" dt 2x ^ zx dt ds dt x dx s dt When the player is 30 ft from first base, x = 60 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 172 Chapter 3 Differentiation 13 and % = -%- (-16) = ^2= « -8.875 ft/sec dt 30a/13 V13 = &■%. Therefore, x = 60 and s = 30x/l3 sx dt ' v (b) cos 0i = ? => -sin^f = -f -| => f -^ (=£-\ = =£ rad/sec; sin 2 ds . dfli _ 90 ds _ _ " " ' dt — sx s 2 sin #i cWi _ 2 = 2p => cos0 2 f dt (30/b) (60) Therefore, x = 60 and s = 30v/l3 => ^ = ^ rad/sec. d«2 _ _ 90 . ds . dfl2_ -90 dt s 2 ' dt dt s 2 cos 9 2 -90 ds dt = ^0 Got) (-15) = (x^floo)f =► x l^ f = I rad/sec = ou ana s = juy u =?• "dt = 65 raa/s \sJ ' \dt) Vs 2 MdJ Vx 2 + 8100-' dt lim 4L x -> dt Cc) dJl _ 90 . ds _ 90 . (x^ v ' dt s 2 sinfli dt ( s2 ';) ^ s ' vat/ \ s* / \ at / \x^ + siuu/ at x — > at „. d02 _ -90 ds _ /"^9o\ /x\ fdx\ _ ( = W\ /dx\ " ' isfl 2 ' dt ~~ \^--J VsJ VdJ ~~ V s 2 ) VdJ -15) = -±rad/sec;f = ? COS #2 38. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and D the distance between the ships. By the Law of Cosines, D 2 = a 2 -f-b 2 2ab cos 120° J(> " 2D [2a|+2b^ + a db dt 2D dt ' " dt b | ] . When a = 5, | = 14, b = 3, and f = 21, then ^ - 4 " 2D where D = 7. The ships are moving ^ = 29.5 knots apart. 3.8 LINEARIZATION AND DIFFERENTIALS 1. f(x) = x 3 - 2x + 3 =>■ f'(x) = 3x 2 - 2 =4> L(x) = f'(2)(x - 2) + f(2) = 10(x - 2) + 7 =>• L(x) = lOx - 13 at x = 2 2. f(x) = Vx 2 + 9 = (x 2 + 9) 1/2 =>• f'(x) = (I) (x 2 + 9) 1/2 (2x) = - | (x + 4) + 5 => L(x) = -|x+|atx=-4 V^ 2 + 9 L(x) = f '(-4)(x + 4) + f(-4) 3. f(x) = x + i => f'(x) = 1 - x- 2 => L(x) = f(l) + f'(l)(x - 1) = 2 + 0(x - 1) = 2 4. f(x) = xV3 ^ f ( X ) = ^L_ => L (x) = f'(-8)(x - (-8)) + f(-8) = i (x + 8) - 2 => L(x) = i x - f 5. f(x) = x 2 + 2x =>■ f'(x) = 2x + 2 =>■ L(x) = f '(0)(x - 0) + f(0) = 2(x - 0) + => L(x) = 2x at x = 6. f(x) = x" 1 =>■ f'(x) = -x" 2 => L(x) = f'(l)(x - 1) + f(l) = (-l)(x - 1) + 1 =4> L(x) = -x + 2 atx = 1 7. f(x) = 2x 2 + 4x - 3 =>• f'(x) = 4x + 4 => L(x) = f'(-l)(x+ 1) + f(-l) = 0(x + 1) + (-5) => L(x) = -5atx= -1 8. f(x) = 1 + x => f (x) = 1 => L(x) = f'(8)(x - 8) + f(8) = l(x - 8) + 9 => L(x) = x + 1 at x = 8 9. f(x) = \/x = x 1 / 3 => f'(x) = (i) x- 2 / 3 => L(x) = f'(8)(x - 8) + f(8) = i (x - 8) + 2 => L(x) =^x+|atx = 8 10. f(x) x+l f'(x) (l)(x+l)-(l)(x) (x+l) 2 (x+l) 2 L(X) = f'(l)(x - 1) + f(l) = 1 (X - 1) + 1 => L(x) =i x +iatx=l Copffigl (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 3.8 Linearization and Differentials 173 11. f(x) = sin x =>■ f (x) = cos x (a) L(x) = f (0)(x - 0) + f(0) = l(x - 0) + => L(x) = x at x = (b) L(X) = f (7T)(X - 7r) + f(7T) = (- 1)(X - 7T) + => L(x) = 7T — xatx = 7r L(x) = TT - X L(x)=x/ " 2 f(x)=sinx 12. f(x) = cos x =^ f'(x) = —sin x (a) L(x) = f (0)(x - 0) + f(0) = 0(x - 0) + 1 => L(x) = 1 at x = (b) L(x) = f'(-f)(x+|)+f(-f) = -(-1) (x + f ) + => L(x) = x + f at x = - 5 2 L(x)=l / ^J-(x)=x+f s ^\ -T, y -\ ' -2 1 s ^2 3 f(x) =C0S X 13. f(x) = sec x =>■ f (x) = sec x tan x (a) L(x) = f (0)(x - 0) + f(0) = 0(x - 0) + 1 => L(x) = 1 at x = (b) L(x) = f (- = -2>/3(x at x = - 5 fM x +i •fi 2 => L(x) = 2 - 2V3 (x + f ) L(x) =2-2/3"(x+i) 0.5 ■*~f (x) = sec x 14. f(x) = tan x => f (x) = sec 2 x (a) L(x) = f (0)(x - 0) + f(0) = l(x - 0) + = x => L(x) = x at x = (b) L(x) = f (|) (x - |) + f (|) = 2 (x - |) + 1 =4> L(x) = 1 + 2 (x ^jatx= % f(x) = tan x X L(x)=l *2(x-J) \k-l 15. f (x) = k(l + x) K \ We have f(0) = 1 and f (0) = k. L(x) = f(0) + f (0)(x - 0) = 1 +k(x - 0) = 1 + kx 16. (a) f(x) = (1 - x) G = [1 + (-x)]° w 1 + 6(-x) = 1 - 6x (b) f(x) = r ^=2[l + (-x)]" 1 «2[l + (-l)(-x)] =2 + 2x (c) f(x) = (l + x)- 1/2 «l+(-i)x=l-| 1 fi (d) f(x) = y/T+** = y/2[ 1+ f ) wy/2(l + If) = 72(1+1 (e) f(x) = (4 + 3x) 1/3 = 4^3(1 + I) 1 / 3 « 4 i/3(i + §f ) = 4^(1 + |) 1X2 2 2 1 :s Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 174 Chapter 3 Differentiation (f) f(x) = (1 - 1 \2/3 2 + x) ' 2 + x, 2/3 3 V 2 + x ) 6+3x 17. (a) (1.0002) 5U = (1 + 0.0002) 5U « 1 + 50(0.0002) = 1 + .01 = 1.01 (b) V 1.009 = (1 + 0.009) 1/3 « 1 + (i) (0.009) = 1 + 0.003 = 1.003 18. f(x) = y/\ + 1 + sin x = (x + l) 1 / 2 + sin x => f (x) = (i) (x + l)" 1 / 2 + cos x => L,(x) = f (0)(x - 0) + f(0) = § (x - 0) + 1 => L, (x) = \ x + 1, the linearization of f(x); g(x) = y/x + 1 = (x + l) 1 / 2 => g'(x) = (i) (x + 1) _1/2 => L g (x) = g'(0)(x - 0) + g(0) = i (x - 0) + 1 =$► L g (x) = i x + 1, the linearization of g(x); h(x) = sin x =^ h'(x) = cos x =^ L h (x) = h'(0)(x - 0) + h(0) = (l)(x - 0) + => L h (x) = x, the linearization of h(x). L f (x) = L g (x) + L h (x) implies that the linearization of a sum is equal to the sum of the linearizations. 19. y = x 3 - 3v/x = x 3 - 3X 1 / 2 => dy = (3x 2 - § x" 1 / 2 ) dx => dy = ^3x 2 - ^A dx 20. y = x V / l-x 2 = x(l-x 2 ) 1/2 => dy= [(l)(l-x 2 ) 1/2 + (x)(±)(l-x 2 ) 1/2 (-2x)j dx 1-x 2 ) 1/2 [(l-x 2 )-x 2 ]dx (l-2x 2 ) Vl-x 2 dx 2x 21. y- — 22. y = H„ _ f (2)(l+x 2 )-(2x)(2x) \ ay ~\ U+x 2 ) 2 J dX =^ dX 2v^ 2xV2 , /x- 1 /2(3(l+x 1 /2))_2x 1 /2(|x- 1 /2)\ ix , _ , , , 3(iT7^ = WTxVT) => dy = ( 9(1 ^ xl/2f J j dx = > , , ox 3X- 1 / 2 + 3-3 9(l+xV2) 2 dy = — ^v^ — f^ dx 3 3^(1 + ^) 2 23. 2y 3 / 2 + xy - x = => 3y 1/2 dy + y dx + x dy - dx = =>■ (3y 1/2 + x) dy = (1 - y) dx => dy ■- 24. xy 2 - 4x 3 / 2 - y = => y 2 dx + 2xy dy - 6X 1 / 2 dx - dy = =4> (2xy - 1) dy = (6X 1 / 2 - y 2 ) dx ^dy=^dx i-y 3^+x dx 25. y = sin (5,/x) = sin (5X 1 / 2 ) => dy = (cos (5X 1 / 2 )) (| x" 1 / 2 ) dx => dy 5 cos (5^/xj dx 26. y = cos (x 2 ) =>■ dy = [—sin (x 2 )] (2x) dx = — 2x sin (x 2 ) dx 27. y = 4 tan Nf\ => dy = 4 (sec 2 (f )) (x 2 ) dx ^ dy = 4x 2 sec 2 (jf) dx 28. y = sec (x 2 - 1) =4> dy = [sec (x 2 - 1) tan (x 2 - 1)] (2x) dx = 2x [sec (x 2 - 1) tan (x 2 - 1)] dx 29. y = 3 esc (l - 2^/x) = 3 esc (l - 2x J / 2 ) => dy = 3 (-esc (l - 2X 1 / 2 )) cot (l - 2X 1 / 2 ) (-x^ 1 / 2 ) dx => dy = 4^ esc (l - 2yfx) cot (l - 2-^x) dx 30. y = 2 cot (-±) = 2 cot (x^ 1 / 2 ) => dy = -2 esc 2 (x^ 1 / 2 ) (- i) (x~ 3 / 2 ) dx =>- dy = -^ esc 2 (-^) 31. f(x) = x 2 + 2x, x = 1, dx = 0.1 => f'(x) = 2x + 2 (a) Af = f(x + dx) - f(x ) = f(l.l) - f(l) = 3.41 - 3 = 0.41 (b) df = f'(x ) dx = [2(1) + 2](0.1) = 0.4 Cop # (c| 1 Pen Educate Inc., publishing as Pearson Addison-Wesle dx Section 3.8 Linearization and Differentials 175 (c) |Af -df I = |0.41 -0.4| =0.01 32. f(x) = 2x 2 + 4x - 3, x = -1, dx = 0.1 => f'(x) = 4x + 4 (a) Af = f(x + dx) - f(x ) = f(-.9) - f(-l) = .02 (b) df=f'(x )dx=[4(-l) + 4](.l) = (c) |Af-df| = |.02-0| = .02 33. f(x) = x 3 - x, x = 1, dx = 0.1 =>■ f'(x) = 3x 2 - 1 (a) Af = f(x + dx) - f(x ) = f(l.l) - f(l) = .231 (b) df=f'(x )dx=[3(l) 2 -l](.l) = .2 (c) |Af-df| = | .231 -.2| = .031 34. f(x) = x 4 , x = 1, dx = 0.1 => f (x) = 4x 3 (a) Af = f(x + dx) - f(x ) = f(l.l) - f(l) = .4641 (b) df = f'fo) dx = 4(l) 3 (.l) = .4 (c) |Af - df | = |.4641 - .4 1 = .0641 35. f(x) = x -1 , x = 0.5, dx = 0.1 =>■ f'(x) = -x~ 2 (a) Af = f(x + dx) - f(x ) = f(.6) - f(.5) = - | (b) df=f'(x )dx = (-4)(i) = -§ (c) |Af-df| = |-| + || = i 36. f(x) = x 3 - 2x + 3, x = 2, dx = 0.1 => f'(x) = 3x 2 - 2 (a) Af = f(x + dx) - f(x ) = f(2.1) - f(2) =1.061 (b) df = f'(x ) dx = (10X0.10) = 1 (c) |Af-df| = 1 1.061 - II = .061 37. V = | ?rr 3 => dV = 47rr 2 dr 38. V = x 3 => dV = 3x 2 dx 39. S = 6x 2 =>• dS = 12x dx 40. S = m v / r y Th 2 = m(r 2 + h 2 ) 1/2 ,hr— - -■ ds - - '- 2 ' ^ 1/2 ^ — - .« j ^-i - dS _ 7T (r 2 + h 2 ) + m 2 dr Vr 2 +h 2 constant =4> ^ = 7r (r 2 + h 2 ) ' + 7rr • r (r 2 + h 2 )~ 2. W+ b2 dS = 7r( ^ + h2) dr, h constant 41. V = 7rr 2 h, height constant =>• dV = 27rroh dr 42. S = 2?rrh => dS = 2?rr dh 43. Given r = 2 m, dr = .02 m (a) A = tit 2 =>■ dA = 2?rr dr = 2tt(2)(.02) = .08tt m 2 (b) (ff)(100%) = 2% 44. C = 27rr and dC = 2 in. = 2tt(5)(±) = 10 in. 2 =^> dC = 2?r dr =>• dr the diameter grew about - in.; A = 7rr 2 =^> dA = 27rr dr 45. The volume of a cylinder is V = 7rr 2 h. When h is held fixed, we have ^ = 27rrh, and so dV = 27rrh dr. For h = 30 in. r = 6 in., and dr = 0.5 in., the volume of the material in the shell is approximately dV = 27rrh dr = 27r(6)(30)(0.5) = 180tt« 565.5 in 3 . Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 176 Chapter 3 Differentiation 46. Let 8 = angle of elevation and h = height of building. Then h = 30tan 6, so dh = 30sec 2 (9 A9. We want |dh| < 0.04h, which gives: |30sec 2 6» d0| < 0.04|30tan6»| ,.fiu \m< 0.04sin |d0| < 0.04sin 6 cos 9 =>• |d0| < 0.04sin ff cos 5jt 12 = 0.01 radian. The angle should be measured with an error of less than 0.01 radian (or approximatley 0.57 degrees), which is a percentage error of approximately 0.76%. 47. V = Trh 3 => dV = 3?rh 2 dh; recall that AV « dV. Then |AV| < (1%)(V) (l)(Th 3 ) 100 IdVl < (l)(vrh 3 ) |37rh 2 dhl < (l)(7rh 3 ) 100 |dh| < joy h = (5 %) h. Therefore the greatest tolerated error in the measurement of his i%. 7T(|) 2 h 7rD?h and h = 10 48. (a) Let D ; represent the inside diameter. Then V = 7rr h — v ,, •■ dV = 57TD; dDi. Recall that AV « dV. We want | AV| < (1%)(V) => |dV| < . 100 J > V 2 ' 5jrD'f ~ 2~ _ 7TD[ ~ 40 5ttD, dD ; < ;vD- dDi < 200. The inside diameter must be measured to within 0.5%. 40 "^ Di (b) Let D e represent the exterior diameter, h the height and S the area of the painted surface. S = 7rD e h dS = TrhdD, dS S ^p . Thus for small changes in exterior diameter, the approximate percentage change in the exterior diameter is equal to the approximate percentage change in the area painted, and to estimate the amount of paint required to within 5%, the tanks's exterior diameter must be measured to within 5%. 49. V = 7rr 2 h, h is constant => dV = 27rrh dr; recall that AV « dV. We want I AVI < I27rrh drl < — =^> Idrl < - L - |z,/uuui| \ 100() =? |ui| \ 2()00 (.05%)r 1000 V IdVl < 7rr 2 h 1000 a .05% variation in the radius can be tolerated. 50. Volume = (x + Ax) 3 3x 2 (Ax) + 3x(Ax) 2 + (Ax) 3 51. W = a+!?=a + bg" 1 => dW = -bg~ 2 dg dW m „„ dW t „ h bdg "(5.2)2 bdg ' "(32)2, 'jk\ = 37.87, so a change of gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth. I try 52. (a) T = 2tt (|) =» dT = 2tt\/l (- \ g" 3 / 2 ) dg = -n^g-V 2 dg (b) If g increases, then dg > =4> dT < 0. The period T decreases and the clock ticks more frequently. Both the pendulum speed and clock speed increase. (c) 0.001 = -TT^/lOO (980~ 3 / 2 ) dg =>• dg w -0.977 cm/sec 2 => the new g w 979 cm/sec 2 53. The error in measurement dx = (1%)(10) = 0.1 cm; V = x 3 => dV = 3x 2 dx = 3(10) 2 (0.1) = 30 cm 3 => the ' 30 \ ,1000 7 percentage error in the volume calculation is (y^) (100%) = 3% Copyright (c) 2006 Pearson Education Section 3.8 Linearization and Differentials 177 54. A = s 2 => dA = 2s ds; recall that AA « dA. Then |AA| < (2%)A m = 51) ^ l dA l - 50 |2sds|<fg Idsl < (2s)(50) 100 (1%) s => the error must be no more than 1% of the true value. 55. Given D = 100 cm, dD = 1 cm, V = \ tt ( D ^ 3 - nti> 3 " V 2 . dV= ^D 2 dD = | (100) 2 (1) 10 4 ji Then ^ (100%) 10 6 ir 6 (10 2 %) infij % = 3% 56. V=|7rr 3 = |7r(§] dV = 2f? dD; recall that AV w dV. Then | AV| < (3%)V 3 ' 100, (*) -D 3 201) |HVl < — => — dD < ^ l uv ! — 200 ^ 2 "^ — 200 dD|<y^ = (l%)D =>■ the allowable percentage error in measuring the diameter is 1%. 57. A 5% error in measuring t => dt = (5%)t = ^ . Then s = 16t 2 =4> ds = 32t dt = 32t (4 32t 2 _ 16t 2 v20; — 20 ~~ 10 .10) = (10%)s =>• a 10% error in the calculation of s. 58. From Example 8 we have y=4y. An increase of 12.5% in r will give a 50% increase in V. 59. lim x^0 '1+x Vl+0 1 + 1 60. lim ^nx x^0 x lim (5&i) (-M „ y Q V X / V COS X / (D(l) = 1 61. E(x) = f(x) - g(x) => E(x) = f(x) - m(x - a) - c. Then E(a) = =>■ f (a) - m(a - a) - c = => c = f(a). Next we calculate m: lim ^ = => lim f(x) ~ m(x ~ a) ~ c = => lim [ f(x) ~ f(a) - ml = (since c = f(a)) x^a x-a x^a x-a x ^ a ^ x -a J v ^ " => f'(a) — m = => m = f '(a). Therefore, g(x) = m(x — a) + c = f '(a)(x — a) + f(a) is the linear approximation, as claimed. 62. (a) i. Q(a) = f(a) implies thatb = f(a). ii. Since Q'(x) = bj + 2b 2 (x - a), Q'(a) = f'(a) implies that bj = f'(a). iii. Since Q"(x) = 2b 2 , Q"(a) = f"(a) implies that b 2 = ^. In summary, b = f(a), bj = f'(a), and b 2 (b) f(x) = (1 - x)- 1 f'(x) = -l(l-x)- 2 (-l) = (l-x)- 2 f"(x) = -2(1 - x)" 3 (-l) = 2(1 - x)~ 3 fja) 2 ' Since f(0) = 1, f'(0) = 1, and f"(0) = 2, the coefficients are b = 1, bj = 1, b 2 approximation is Q(x) = 1 + x + x 2 . 1 . The quadratic (c) As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. [-2.35, 2.35]by [-1.25, 3.25] (d) g(x) = x- 1 g'(x) = -lx- 2 g"(x) = 2x~ 3 Since g(l) = 1, g'(l) = — 1, and g"(l) = 2 , the coefficients are bo = 1, bi -l,b 2 1 . The quadratic Cfiojt (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 178 Chapter 3 Differentiation approximation is Q(x) = 1 — (x — 1) + (x — 1) As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. [-1.35,3.35]by[-1.25,3.25] (e) h(x) = (l+x) 1/2 h'(x) = i(l + x)- 1/2 h"(x) = -i(l+x)- 3 / 2 Since h(0) = 1, h'(0) = \, and h"(0) approximation is Q(x) = 1 + | — y. -\ , the coefficients are bo = 1, bi = \,^i = -f = — |. The quadratic As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. [-1.35, 3.35]by [-1.25, 3.25] (f) The linearization of any differentiable function u(x) at x = a is L(x) = u(a) + u'(a)(x — a) = bo + bi(x — a), where bo and bi are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization for f(x) at x = is 1 + x; the linearization for g(x) at x = 1 is 1 — (x — 1) or 2 — x; and the linearization for h(x) at x = 0isl + |. 63. (a) x = 1 5rr2 Y-*y\ 9 0. 95 j* l.jOS 1 1 i (b) x= l;m = 2.5, e 1 « 2.7 9 0. 95 1. *S 1 y.S/ frrS x = 0;m= l,e°= 1 -*T«i «.l -0-05 -«;-9 05 l i Oil l;m = 0.3, e- 1 w 0.4 Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Section 3.8 Linearization and Differentials 179 64. If f has a horizontal tangent at x = a, then f ' (a) = and the linearization of f at x = a is L(x) = f(a) + f'(a)(x — a) = f(a) + - (x — a) = f(a). The linearization is a constant. 65. Find |v| when m = l.Olnio. m = => |v| = c\/l - 55? => dv = c • dv= - t^T^ m A /l 2\ x n?) V m3 / m = ^/I - K = ^ =■ 1 dm, dm = O.Olmo =>• dv = v 2 _ m 2 c 2 m 2 mp m 2 VlOO^ 1-1 c-m* 101 3 m 3 1— #- \j 100 2 o 'mo.) -100) 1000 101 3 i/l- 100 2 10 1 2 0.69c. Body at rest => v = and v = v + dv 0.69c. c* 1 „, — 101 ^ m - M m , " 2 / 66. (a) The successive square roots of 2 appear to converge to the number 1 . For tenth roots the convergence is more rapid, (b) Successive square roots of 0.5 also converge to 1. In fact, successive square roots of any positive number converge tol. A graph indicates what is going on: p- x 0.5 1 1.5 2 Starting on the line y = x, the successive square roots are found by moving to the graph of y the line y = x again. From any positive starting value x, the iterates converge to 1 . ^/x and then across to 67-70. Example CAS commands: Maple : with(plots): a: = l:f:=x->xA3+A- 2 2*x; plot(f(x), x=-1..2); diff(f(x),x); fp := unapply (",x); L:=x -> f(a) + fp(a)*(x - a); plot({f(x),L(x)),x=-1..2); err:=x -> abs(f(x) — L(x)); plot(err(x), x=— 1..2, title = #absolute error function*); err(— 1); Mathematica : (function, xl, x2, and a may vary): Clearff, x] {xl,x2} = {-l,2};a = 1; f[x_]:=x _ v 3 2x Plot[f[x],{x,xl,x2}] lin[x_]=f[a] + f [a](x - a) Plot[{f[x],lin[x]},{x,xl,x2}] err[x_]=Abs[f[x] - lin[x]] Copyright (c) 1 Pearson E Won, k, publishing as Pearson Addison-Wesle 180 Chapter 3 Differentiation Plot[err[x],{x,xl,x2}] err//N After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and delta (del) eps = 0.5;del = 0.4 Plot[{err[x], eps},{x, a — del, a + del}] CHAPTER 3 PRACTICE EXERCISES 1. y = x 5 - 0.125x 2 + 0.25x =>• f = 5x 4 - 0.25x + 0.25 2. y = 3 - 0.7x 3 + 0.3x 7 => 3. y = X 3 - 3 (X 2 + 7T 2 ) => 4. y = x 7 + Jlx - g = -2.1x 2 + 2.1x 6 | = 3x 2 - 3(2x + 0) = 3x 2 - 6x = 3x(x - 2) ^T =* %=T? + S> 5. y = (x+ l) 2 (x 2 + 2x) = 2(x+ l)(2x 2 + 4x+ 1) dv (x + l) 2 (2x + 2) + (x 2 + 2x) (2(x + 1)) = 2(x + 1) [(x + l) 2 + x(x + 2)] 6. y = (2x - 5)(4 - x)" 1 => g = (2x - 5)(-l)(4 - x)- 2 (-l) + (4 - x)- x (2) = (4 - x)- 2 [(2x - 5) + 2(4 - x) = 3(4 - x)-' 2 7- y 9 2 + sec 9 + l) 3 => t = 3 (6» 2 + sec 9 + \f{26 + sec 9 tan 0) 8. y (_, _ cscj? _ fpY v 2 " v -■ ■ - 2 (- J> -)( 5 CSC ( 2 | (esc 61 cot 9 - 6) 9. s Vt ^ ds _ o+v^-^-^^) _ (i + v^)-yt i+Vt dt (i+^t) 20(1 + 7^ 20(1 + ^ 10. s i ^ ds= (>A-0(Q)-i(^) = _i v/t-l dt (v^-1) 2 2^(^-l) 2 11. y = 2 tan 2 x — sec 2 x =>• gf = (4 tan x) (sec 2 x) — (2 sec x)(sec x tan x) = 2 sec 2 x tan x 12. y = jjp-j — -tj- = esc 2 x — 2 esc x =>• ■£ = (2 esc x)(— esc x cot x) — 2(— esc x cot x) = (2 esc x cot x)(l — esc x) 13. s = cos 4 (1 - 2t) => | = 4 cos 3 (1 - 2t)(-sin (1 - 2t))(-2) = 8 cos 3 (1 - 2t) sin (1 - 2t) 14. s = cot 3 (|) * | = 3 cot 2 (|) (-esc 2 (f)) (f ) = | cot 2 (f) esc 2 (f) 15. s = (sec t + tan t) 5 => ^ = 5(sec t + tan t) 4 (sec t tan t + sec 2 t) = 5(sec t)(sec t + tan t) 5 16. s = csc 5 (l -t + 3t 2 ) =4> f t =5csc 4 (l -t+3t 2 )(-csc(l -t + 3t 2 )cot(l - t + 3t 2 )) (-1 + 6t) = -5(6t- l)csc 5 (l -t + 3t 2 )cot(l — t + 3t 2 ) 17. r = JlB sin e = (29 sin (9) 1/2 => ^ = ± (26> sin 9)- l l 2 (29 cos 6> + 2 sin 0) = gco / se + sing ay l y 20 sin Copfiigl (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Chapter 3 Practice Exercises 181 18. r = 26> a/cos 9 = 29 (cos 9) 1 !' 2 _ 2 cos 9-9 sin 9 29 (1) (cos 6»)- 1 / 2 (-sin (9) + 2(cos 6>) 1/2 = =p^ + 2^fcos9 /cos 9 19. r = sin ^29 = sin(20) 1/2 => % = cos(29) 1 / 2 (1 (26T 1/2 (2)) = 52^ 20. r = sin (0 + v^ (. + ^TT)(i + ^ TT ) = 2 ^cos(. + ^TT) i v2 re ? -^ dy - I v2/_ m , 2 „- t 2\ /=2 21. y = 5X 2 esc = =>• x — esc £ cot 2\ i\ X X / V x^ (csc -I I $ ■ 2xj = csc - cot - + x csc 22. y = 2v/x" sin yfi => g = 2^ (cos ^ ( 0-) + (sin ^/x) (^) = cos y/i 2\A 2 \ / — , sin ~ ! — COS \/X sA y^ 23. y = x- 1 / 2 sec (2x) 2 J) - 1 / 2 sec (2x) 2 tan(2x) 2 (2(2x) • 2) + sec (2x) 2 (- ± x" 3 / 2 ) = 8X 1 / 2 sec(2x) 2 tan(2x) 2 - \ x~ 3 / 2 sec(2x) 2 = \ x 1 / 2 sec(2x) 2 [16 tan(2x) 2 - x~ 2 ] or ^sec(2x) 2 [l6x 2 tan(2x) 2 - l] 24. y = v/x csc (x + l) 3 = x 1 / 2 csc (x + l) 3 (-csc(x+ l) 3 cot(x+ l) 3 )(3(x+ l) 2 )+csc(x+ l) 3 (ix^ 1 /' 2 ) dy _ Y l/2 dx — A -3v/x(x+ l) 2 csc(x+ l) 3 cot(x+ l) 3 + csc ^ x + 1)3 = \ v /xcsc(x+ l) 3 [i - 6(x+ l) 2 cot(x+ l) 3 ] or A^csc(x + 1) 3 [1 - 6x(x + l) 2 cot(x + l) 3 ] 25. y = 5cotx 2 => ^ = 5 (-csc 2 x 2 ) (2x) = -lOx csc 2 (x 2 ) 26. y = x 2 cot 5x =$■ g| = x 2 (-csc 2 5x) (5) + (cot 5x)(2x) = -5x 2 csc 2 5x + 2x cot 5x 27. y = x 2 sin 2 (2x 2 ) => g = x 2 (2 sin (2x 2 )) (cos (2x 2 )) (4x) + sin 2 (2x 2 ) (2x) = 8x 3 sin (2x 2 ) cos (2x 2 ) + 2x sin 2 (2x 2 ) 28. y = x~ 2 sin 2 (x 3 ) =4> & = x~ 2 (2 sin (x 3 )) (cos (x 3 )) (3x 2 ) + sin 2 (x 3 ) (-2x~ 3 ) = 6 sin (x 3 ) cos (x 3 ) - 2x~ 3 sin 2 (x 3 ) on „ _ (JL_\- 2 -v ds _ o (^\- 3 f (t+l)(4)-(4t)(l) \ _ _ 9 /Jl\-3 4 _ _ (t+1) Zy - S— Vt+lJ ^ dt ~~ Z Vt+lJ ^ (t+1) 2 J — z Vt+lJ (t+1) 2 — 8t? 30. s -l 15(15t-l) 3 i (15t - I)" 3 =► | = - i (-3)(15t - 1) 4 (15) = ^ I)" 1 3i-y = (x4) =*! = 2(i&) vM , »+')(^)~(>A)(') _ (x+1)- 2x 1 - x (x+1) 2 (X+1)3 (X+1)3 32 -y = {^) : Wi \ / fiyg+i^-fty^) ^ _ Vx(4 E ) dy _ 2 f 2 \A A dx V2V^+l/ I (2,/x+l) 2 (2^+l) J (2^+l) J 33. y x 2 + x X 2 IN 1/2 dy -1/2 2x 2 \\ + I 34. y = 4x^x + ^ = 4x (x + x 1 ^) 1/2 ^ g = 4 x (I) (x + x 1 ^) ~ 1/2 (i + i x -i/2) + ( x + x i/2) Va (4) = (x +v ^)- 1/2 [2x(l + ^) + 4(x +v ^)]=(x +v ^)- 1/2 (2x + ^ + 4x + 4^) = ^l| Copy fight (c| 1 Pearson Etati, Inc., publishing as Pearson Addison-Wesle 182 Chapter 3 Differentiation 35. r: V cos 8 - 1 / dr --> / sin t d0 ~ Z V cos 9 - 1 / 9 \ (cos - l)(cos 0) - (sin 9)(-sin t (cos 9- l) 2 f sin 8 \ ( cos 2 3 - cos 8 + sin 2 8 \ Vcos - 1 ) \ (cos 9 - l) 2 J (2 sin 9) (1 -cost -2 sin 8 (cos9-l) 3 (cosfl-1) 2 36. r ' sin 8 + 1 ' .l-cosfly " dfl ^Vl-cosfl; | (l-cos9) 2 2(sin 9 + 1) / a „_„2 fl c ;„2 /) „;_, a\ _ 2(sin 9+ l)(cos 9-sin 9- 1) dr _ o /• sin 8 + \\ (1 - cos 9)(cos 0) - (sin 9 + l)(sin t (l-cos9) 3 (cos — cos 2 — sirr — sin 0) (1 -cos9) 3 37. y = (2x + 1) ^2x4-1 = (2x + l) 3 / 2 =}► g = § (2x + l) 1 / 2 (2) = 3^2x+l 38. y = 20(3x - 4) 1 / 4 (3x - 4)- 1 / 5 = 20(3x - 4) 1 / 20 =>■ g = 20 (i) (3x - 4)- 19 / 20 (3) 3 (3x-4)i9/ 2 o 39. y = 3 (5x 2 + sin 2x)- 3/2 => g = 3 (- |) (5x 2 + sin 2x)- 5/2 [10x + (cos 2x)(2)] = g^ffl 40. y = (3 + cos 3 3x) 1/3 => g = - i (3 + cos 3 3x) 4/3 (3 cos 2 3x) (-sin 3x)(3) = ^J™ 2 3 * fj% [j ~j~ COS jXI 41. xy + 2x + 3y = 1 => (xy' + y) + 2 + 3y' = => xy' + 3y' = -2 - y => y'(x + 3) = -2 - y => y' = - |±| 42. x 2 +xy + y 2 -5x = 2 => 2x + (x g + y) + 2y g - 5 = =* xg + 2yg = 5-2x-y =* g (x + 2y) = 5-2x-y^ g = ^ 43. x 3 + 4xy - 3y 4 / 3 = 2x => 3x 2 + (4x g + 4y) - 4y x / 3 g = 2 =► 4x | - 4y 1 / 3 | = 2 - 3x 2 - 4y => S(4x-4yV3) =2 -3x 2 -4y^ g = *^£ 44. 5x 4 / 5 + 10y 6 / 5 = 15 => 4X- 1 / 5 + ^y 1 / 5 g = =>■ ^y 1 / 5 g = -4X- 1 / 5 =>■ g = - i x -i/5 y -i/5 = _ _1_ 45. (xy) 1 ^ = 1 => ^(xy)- 1 ^ ( x g + y ) = => x^y-l^ g = _ x -i/2 y i/2 /2 _^ dy dx -^y => d I = -l 46. x 2 y 2 = 1 => x 2 (2y g) + y 2 (2x) = => 2x 2 y g = -2xy 2 => g x+l 2 dy _ (x+l)(l)-(x)(D _^ dy _ 1 ^ dx 47. y 2 - ■ K 48. y >=(H)''-' =-y' = j^ =-4v iK l+x\l/2 (x+l) 2 ' ~~^ dx 2y(x+l) 2 3 dy _ (l-x)(l)-(l+x)(-l) dy (1-x) 2 dx ~~ 2y 3 (l-x) 2 49. p 3 + 4pq-3q 2 = 2 => 3p 2 g + 4 (p + q g) - 6q = =4- 3p 2 g + 4q g = 6q - 4p dp 6q — 4p dq ~~ 3p 2 + 4q | (3p 2 + 4q) = 6q - 4p 50. q = (5p 2 + 2p)- 3/2 => 1 = - | (5p 2 + 2p)- 5/2 (lOp | + 2 g) => - f (5p 2 + 2p) 5/2 = g (10p + 2) dp (5p 2 + 2p)° /2 ^ dq 3(5p+l) Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Chapter 3 Practice Exercises 183 51. r cos 2s + sin 2 s = tt =>• r(— sin 2s)(2) + (cos 2s) (^) + 2 sin s cos s = => ^ (cos 2s) = 2r sin 2s — 2 sin s cos s ^ dr = 2r sin 2s -sin 2s = (2r- l)(sin 2s) = (2 _ j )( &) as cos 2s cos 2s v ' v / 52. 2rs - r - s + s 2 = -3 =^ 2 (r + s |) - % - 1 + 2s = => f s (2s - 1) = 1 - 2s - 2r => ^ dr 1 - 2s - 2r 2s- 1 53. (a) x 3 + y 3 = 1 => 3x 2 + 3y 2 g = => g - x , - d 2 y _ -2xy 2 + (2yx 2 )(-^) _ -2xy 2 - ^ _ _ 2xy 3 _ 2x 4 x 2 _ d 2 y _ y 2 (-2x)-(-^)(2yg) d\- dx 2 ,2 _ r 2 (b) y 2 =l-f =»2y£=£ =* 1 = ^ => d | = (yx 2 r 1 => g = -(y X 2 r 2 fy(2x) + x 2 d | d 2 y dx 2 — 2xy — x 2 I ^2 -2xy 2 - 1 y^x*> 54. (a) x 2 - y 2 = 1 => 2x - 2y g = => -2y g = -2x - ^ (b) dZ _ x . d2y = y(D-*5* dx y dx 2 y 2 y- x u v 2 -^ 2 dx y 2 v 2 *^ = 4 (since y 2 -x 2 =-l) 55. (a) Let h(x) = 6f(x) - g(x) => h'(x) = 6f (x) - g'(x) => h'(l) = 6f (1) - g'(l) = 6 (§) - (-4) = 7 (b) Let h(x) = f(x)g 2 (x) =► h'(x) = f(x) (2g(x)) g'(x) + g 2 (x)f (x) => h'(0) = 2f(0)g(0)g'(0) + g 2 (0)f (0) = 2(l)(l)(i)+(l) 2 (-3)=-2 (c) Leth(x) f(x) (x)+l (5 + !)(!) -3 (-4) _ 5_ (5+1) 2 12 h'(x) (g(x)+l)f'(x)-f(»)g'W (g(x)+l) 2 h'(l) (g(l) + l)f'(l)-f(l)g'(l) (g(l)+l) 2 (d) Leth(x) = f(g(x)) => h'(x) = f (g(x))g'(x) =>• h'(0) = f'(g(0))g'(0) = f'(l) (|) = (i) (|) = f (e) Leth(x) = g(f(x)) => h'(x) = g'(f(x))f (x) =* h'(0) = g'(f(0))f (0) = g'(l)f (0) = (-4) (-3) = 12 (f) Let h(x) = (x + f(x)) 3 / 2 =>• h'(x) = § (x + f(x)) : / 2 (1 + f (x)) =^> h'(l) = § (1 + f(l)) 1/2 (1 + f'(l)) = 1(1 + 3)V 2 (1 + I) = | (g) Let h(x) = f(x + g(x)) =► h'(x) = f'(x + g(x)) (1 + g'(x)) =► h'(0) = f'(g(0)) (1 + g'(0)) f'(l)(l + i) = (|) (|) 56. (a) Let h(x) = ^f(x) =» h'(x) = ^f'(x) + f(x) • ^ => h'(l) = >/l f (1) + f(l) • A- = ± + (-3) (i 2\/T B 10 (b) Leth(x) = (f(x))V 2 => h'(x) = \ (f(x))~V2 (f ( x) ) => h'(0) = | (f(0))- x / 2 f'(0) = \ (9)-^ 2 (-2) = - | (c) Leth(x) =f (y^) => h'(x) = f (-y^) " 3^ => h'(l) = f (v^) ■ ^j = H = To (d) Let h(x) = f(l - 5 tan x) => h'(x) = f'(l - 5 tan x) (-5 sec 2 x) =4> h'(0) = f'(l - 5 tan 0) (-5 sec 2 0) = f'(l)(-5)=±(-5) = -l (e) Leth(x) = =JE- ^ h ' (x) = (2 + cosx)fV)-f(x)(-sinx) h , (Q) = (2 + l)f'(0)-f(0)(0) = 3(_2) = _ 2 v y v ' 2 + cosx v y (2 + cosx)^ 7 (2+iy 9 3 (f) Leth(x)= 10sin(f)f 2 (x) => h'(x) = 10 sin (f ) (2f(x)f (x)) + f 2 (x) (10 cos (f )) (f ) => h'(l) = 10 sin (|) (2f(l)f'(l)) + f 2 (1) (10 cos (f )) (f ) = 20(-3) (i) + = -12 57. x = t 2 + tt => ^ = 2t; y = 3 sin 2x => g = 3(cos 2x)(2) = 6 cos 2x = 6 cos (2t 2 + 2tt) = 6 cos (2t 2 ) ; thus, f = g-f =6cos(2t 2 )-2t => %\ =6cos(0)-0 = I t=0 58. t = (u 2 + 2u) 1/3 =>• £ = 5 (u 2 + 2u)~ 2/3 (2u + 2) = | (u 2 + 2u)~ 2/3 (u + 1); s = t 2 + 5t =>■ | = 2t + 5 = 2 (u 2 + 2u) 1/3 + 5; thus | = * • | = [2 (u 2 + 2u) 1/3 + 5] (§) (u 2 + 2u)~ 2/3 (u + 1) Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 184 Chapter 3 Differentiation ds I du I u=2 [2 (2 2 + 2(2)) 1/3 + 5] (|) (2 2 + 2(2)r 2/3 (2 + 1) = 2 (2 - 8 1 / 3 + 5) (S" 2 / 3 ) =2(2-2 + 5) (1) = | 59. r = 8 sin (s dr ds cos (s + |J ; w = sin (aA-2) COS I l) /8sin(s + g)-2 ;thu ^^ Ja Jr a,UN-n|s f| 2! ; (7* 2*/8sin (s- dw I ds I s=o ds dr ds 2^8sin(s + g) [8cos(s+|)] . (^8 sin (I) -2) -8 cos (f) (cos0)(8)(^) 2j8sin(f) 2x/4 V~3 60. dh + 9 = 1 => (<9 2 + t (26> f )) + f = 1 (0 2 + 7)~ 2/3 (20) and dr I I 9=1 ~~ 3 (1+7) -2/3 dr I _ dr I _ dfl I dt I t=0 A6 I 1=0 dt I t=0 f (20t + 1) = -<9 2 d0 dt 20t+l ; r 7) 2/3 ;nowt = 0and(9 2 t + 6> = 1 (c) 2 + 7) 1/3 1 so that dt I t=o. 9=1 1 61. y + y = 2 cos x -2sin(0) 3y 2 ^ dx dy dx -2 sin x l(3y 2 i) -2 sin x ill dx -2 s 3y 2 +l d 2 y _ (3y 2 +l)(-2cosx)-(-2 S inx)(6y£) 0; dx 2 3+1 "' dx 2 (3y 2 +l) 2 d 2 y I _ (3 + l)(-2 cos 0) - (-2 sin 0)(6-0) dx 2 I , - (3 + 1)2 dy dx (0,1) 62. x 1 / 3 + y 1 / 3 = 4 => i x-' 2 / 3 + i y- 2 / 3 S = => £ v 2/3 dy dx 2/3 1 . dy _ _ x ' dx — x 2 / 3 d 2 y = (* 2/3 )(-ty- 1/3 S)-(-y 2/3 )(!*- 1/3 ) dx 2 ( x 2/3) 2 d 2 ^ dx 2 (S.Sl (8 2 -' 3 ) [- |-8-V 3 -(-l)] + (8 2 / 3 ) (1-8-V 3 ) 8"/ 3 8 2 3 63. f(t) = jrh - and f ( l + h ) 2(t + h)+l -2 1 1 f(t + h) - f(t) _ 2(t+h)+i 2t+i _ 2t+l-(2t + 2h+l) h h ~ (2t + 2h+l)(2t+l)h -2 (2t + 2h+l)(2t+l)h (2t + 2h+l)(2t+l) -2 (2t+l) 2 f( t ) = lim m + h)-nt) W h->0 h h-?U (2t + 2h+l)(2t+l) 64. g(x) = 2x 2 + 1 and g(x + h) = 2(x + h) 2 + 1 = 2x 2 + 4xh + 2h 2 + 1 = (2x 2 + 4xh + 2h 2 i +l)-(2x 2 +l) = 4xhi2h! = 4x + 2h ^ g , (x) = ^ , g(x + h) - g(x) ^ h g(x + h)-g(x) h lim (4x + 2h) h-»0 4x 65. (a) v > .V -1 \ l -l - \ /w = x z ,-lS;t<0 (b) lim f(x) = lim x 2 = and lim f(x) = lim -x 2 = => lim f(x) = 0. Since lim f(x) = = f(0) it x -> 0" x -» (T x -> 0+ x -> 0+ x -> x -» follows that f is continuous at x = 0. (c) lim f ' (x) = lim (2x) = and lim f ' (x) = lim (— 2x) = =>- lim f (x) = 0. Since this limit exists, it x -> 0" x -> 0" x -> 0+ x -> 0+ x -> Cfiojt (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Chapter 3 Practice Exercises 185 follows that f is differentiable at x = 0. 66. (a) '<*) = f x, -1Sx<0 itanx, OsxSn/4 it/4 (b) lim f(x) = lim x = and lim f(x) = lim tan x = => lim f(x) = 0. Since lim f(x) = = f(0), it x^O" x^O" x ^ 0+ x^0 + x^O x^O follows that f is continuous at x = 0. (c) lim f'(x) = lim 1 = 1 and lim f'(x) = lim sec 2 x=l => lim f'(x) = 1. Since this limit exists it x^O" x^O" x ^ 0+ x ^ 0+ x^O follows that f is differentiable at x = 0. 67. (a) x, 0<x<] 2-x, 1 <x<2 (b) lim f(x) = lim x = 1 and lim f(x) = lim (2 — x) = 1 x->l" x->l- x^l+ x^l+ follows that f is continuous at x = 1. (c) lim_ f'(x) = lim 1 = 1 and lim f'(x) = lim -1 = -1 x —> 1 x^l x ^ 1+ x ^ 1+ not exist => f is not differentiable at x = 1 . lim f(x) = 1. Since lim f(x) = 1 = f(l), it X —> 1 X — > 1 lim_ f'(x) y^ lim f'(x), so lim f'(x) does x^l x ^ 1+ x — > 1 68. (a) lim f(x) = lim sin 2x = and lim f(x) = lim mx = => lim f(x) = 0, independent of m; since x^O" x^CT x ^o + x^O + x^O f(0) = = lim f(x) it follows that f is continuous at x = for all values of m. x — > (b) lim f'(x) = lim (sin 2x)' = lim 2 cos 2x = 2 and lim f'(x) = lim (mx)' = lim m = m =£- fis x -> 0" x -> 0" x -> 0" x -> 0+ x -> 0+ x -> 0+ differentiable at x = provided that lim f'(x) = lim f'(x) =>• m = 2. x -> x -> 0+ 69. y = f + 2^4 = 5 x + (2x - 4)" 1 => g| = \ - 2(2x - 4)~ 2 ; the slope of the tangent is — § => - § = i - 2(2x - 4y 2 => -2 = -2(2x - 4)~ 2 => 1 = ^^ => (2x - 4) 2 = 1 =>■ 4x 2 - 16x + 16 = 1 => 4x 2 - 16x + 15 = =4> (2x - 5)(2x - 3) = =>■ x = | or x = | => (|, |) and (§, - ±) are points on the curve where the slope is — | . 70. y = x-i => | = i + ^ => x = ±5 => (i and 1 + i ; the slope of the tangent is 3 => 3 = 1 + ^3 =>• 2=^ — 5, 5) are points on the curve where the slope is 3. 71. y = 2x 3 — 3x 2 — 12x + 20 =>• ■£ = 6x 2 — 6x — 12; the tangent is parallel to the x-axis when ^ = =>■ 6x 2 - 6x - 12 = => x 2 - x - 2 = => (x - 2)(x + 1) = => x = 2 or x = -1 => (2, 0) and (-1, 27) are points on the curve where the tangent is parallel to the x-axis. 72. y = x 3 => dy 3x 2 dy dx 12; an equation of the tangent line at (—2, —8) is y + 8 = 12(x + 2) Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 186 Chapter 3 Differentiation => y= 12x + 16; x-intercept: = 12x + 16 |, 0) ; y-intercept: y = 12(0) + 16 = 16 =» (0, 16) 73. y = 2x 3 - 3x 2 - 12x + 20 => & = 6x 2 - 6x - 12 (a) The tangent is perpendicular to the line y = 1 — |j when -Z (=to) 24; 6x 2 - 6x - 12 = 24 =>• x 2 - x - 2 = 4 =4> x 2 -x-6 = =4> (x- 3)(x + 2) = =S> x=-2orx = 3 => (-2, 16) and (3, 1 1) are points where the tangent is perpendicular to y = 1 — ^ . (b) The tangent is parallel to the line y = \/l - 12x when ^ = -12 => 6x 2 - 6x - 12 = -12 => x 2 - x = =>• x(x — 1) = =>■ x = or x = 1 =>• (0, 20) and (1, 7) are points where the tangent is parallel to y = \fl - 12x. 74 v = Tsmx . * x dy x(7r cos x) — (tt sin x)(l) dx x 2 mi dy dx tt2 Since nij = — -i- the tangents intersect at right angles. 1 and m 2 dy dx I 7T 2 I X-— 7T 75. tan x, f <*<f dy dx sec 2 x; now the slope of y = — | is — i =>• the normal line is parallel to | when -# 2 dx 9 1 ,S X = 2 = 2. Thus, sec J x ±i v/2 | and x for - | < x < | => |, — l) and (f ( l) are points where the normal is parallel to y = — | . 76. y = 1 dy dx dy dx (1.0 the tangent at (|, l) is the line y — 1 = — (x — |) y = —x + | + 1; the normal at l = (l)(x-f) => y = x-f 1 is 77. y = x 2 thus, 1 C :§) 2 *■ = 2x and y = x C => C=i dx 1 ; the parabola is tangent to y = x when 2x = 1 1 . 2 - dy dx 78- y = x 3 =► I = 3x 2 =■ ;t intersects y = x 3 when x 3 => (x - a) 2 (x + 2a) = = 3a 2 => the tangent line at (a, a 3 ) is y — a 3 = 3a 2 (x — a). The tangent line 3a 2 (x - a) => (x - a) (x 2 + xa + a 2 ) = 3a 2 (x - a) => (x - a) (x 2 + xa - 2a 2 ) = = a or x = -2a. Now p- 1 = 3(-2a) 2 = 12a 2 = 4 (3a 2 ), so the slope at x = —2a is 4 times as large as the slope at (a, a 3 ) where x = a. 3 -(-2) 79. The line through (0,3) and (5, —2) has slope m — (| _ - y = -x + 3; y = ^fj => ^ = (x ~ c 1)2 , so the curve is tangent to y = -x + 3 => (x+l) 2 =c,x/-l. Moreover, y = ^-y intersects y = — x + 3 =^> ^ 1 => the line through (0,3) and (5, -2) is ^z — _i dx ~~ L (1+1)2 x+3,x/ -1 => c = (x + l)(-x + 3), x ^ -1. Thus c = c =>• (x + l) 2 = (x + l)(-x + 3) => (x + l)[x + 1 - (-x + 3)] = 0, x ^ -1 => (x + l)(2x - 2) = =>■ x = 1 (since x ^ -1) => c = 4. Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Chapter 3 Practice Exercises 187 80. Let ( b, ± v a 2 — b 2 j be a point on the circle x 2 + y 2 = a 2 . Then x dy dx ±\/a 2 -b 2 normal line through (b, ± \J a 2 — b 2 J has slope - 2 ^2x + 2yg=0=> g ± v' d - h _^ normal line is y - ( ± \/a 2 - b 2 ) = ± v / ^ (x - b) =» y T \/a 2 - b 2 = ± v / ^ x T \A 2 - b 2 => y = ± ^^ x which passes through the origin. $1. x 2 + 2y 2 = 9 =>• 2x + 4y ^ =- () dx dy dx 2y dy dx | and the normal line is y = 2 + 4(x — 1) = 4x — 2. the tangent line isy = 2— j(x— 1) 82. x 3 + y 2 = 2 => 3x 2 + 2y ^ = dy _ -3x : dx 2y dy dx -fx- (1,1) 2 A, ix _ 2„ and the normal line is y = 1 + 4 (x — 1) the tangent line isy= 1 + ^ (x — 1) 83. xy + 2x - 5y = 2 =* (x | + y) + 2 - 5 g = =*■ g (x - 5) = -y - 2 =>■ dy dx -y-2 x-5 dy dx (3,2) the tangent line is y = 2 + 2(x — 3) = 2x — 4 and the normal line isy = 2+ -J- (x — 3) = — lx+ I . 84. (y - x) 2 = 2x + 4 => 2(y - x) (g - l) = 2 => (y - x) g = 1 + (y - x) =► dy 1 + y — x dx y — x dy dx (6,2) the tangent line is y = 2 + | (x — 6) = |x- | and the normal line is y = 2 — I (x — 6) = — | x + 10. 85. x xy = 6 1 *?(«£+*) o dy + y = -2,/xy =>• dy _ -2x/xy-y dy dx the tangent line isy^ 1 — |(x — 4) = — |x + 6 and the normal line isy=l+^(x — 4)=^x— ^. 86. x 3 / 2 + 2y 3 / 2 = 17 => § x 1 / 2 + 3y J / 2 g = dy -xV2 dx 2yV2 dy dx the tangent line is y = 4-l(x-l) ■^ and the normal line is y = 4 + 4(x — 1) = 4x. 87. x 3 y 3 + y 2 = x + y => [x 3 ( 3 y 2 g) + y 3 (3x 2 )] + 2y g = 1 + g => 3x 3 y 2 g + 2y g - g = 1 - 3x 2 y 3 - ^vV + 2>-l) = l-AV - | = 3^f^T •-- ; - ^ ■ '-■ ix ' ? ' - S =-|,but£ is undefined. l(i,D ax l(i-D Therefore, the curve has slope — | at (1, 1) but the slope is undefined at (1, — 1). y = sin (x — sin x) =>• -^ = [cos (x — sin x)](l — cos x); y = =>• sin (x — sin x) = =4> x — sin x = kn, k = —2, —1,0, 1,2 (for our interval) => cos (x — sin x) = cos (k7r) = ± 1. Therefore, -Z = and y = when 1 — cos x = and x = k7r. For — 2n < x < 27r, these equations hold when k = —2, 0, and 2 (since cos (—7r) = cos 7r = —1). Thus the curve has horizontal tangents at the x-axis for the x-values — 2ir, 0, and 2ir (which are even integer multiples of n) =>• the curve has an infinite number of horizontal tangents. x = = tan t, y = = sec t =► £ dy _ dy/dt _ \ sec t tan t _ tan t dx dx/dt \ sec 2 1 sin t dy dx sin vr _ V3. t _ t=x/3 =>■ x = \ tan | = = 2cos 3 (f)=I 90. x = l+i,y = l f andy=Isecf = l => y = f x + \ ; g = ^ = ^ 2cos 3 t - ^ dx 2 I i =7r /3 dy dx dy/dt _ (?) dx/dt ~~ I 2 dv dx (2) = -3; t = 2 => x = 1 22 and Cop # (c) 1 Pearson Education, Ire,, publishing as Pearson Addison-Wesle 188 Chapter 3 Differentiation y = l- -3x 13 . _y _ dyTdt _ (- f) 4 ' dx 2 dx/dt / 2\ l* ^ - ^ (2) 3 = 6 91. B = graph of f, A = graph of f. Curve B cannot be the derivative of A because A has only negative slopes while some of B's values are positive. 92. A = graph of f, B = graph of f '. Curve A cannot be the derivative of B because B has only negative slopes while A has positive values for x > 0. 93. (-1, 2) 94. H.0) y = f(x) (t.1) ZX (6.-1) (1.-2) 95. (a) 0,0 (b) largest 1700, smallest about 1400 96. rabbits/day and foxes/day 97. lim x^0 2x2 -X lim x^ ' sin x \ 1 x ) (2x-l) (D(i)=-1 98. lim x^0 3x — tan 7x lim x^0 (|5 - ^V) = \ - lim (-Kr ■ s^5 . i) = 3(1.1.7) V2x 2xcos7x/ 2 x — * \ cos ^ x 7x (I) / * ^ 99. lim r-»0 tan 2r ._?„ (^ • SHI? - I) = (I) CD l™ fipT = (5) (D (l) = 5 r^0 b (¥) 0^0 lim -> sm fl 0^0 sin (sin 8) .._, sin x / sin (sin g) \ / sin g \ V_ sinfl ) \ 6 I 100. lim 5»(™9) = lim [m^l\ (»infl) = i im S^E> . L etx = sinfl. Then x -» as ->0 lim x^0 (9^0 101. lim 4tan^ + t^ +1 fl i % \ - tan 2 9 + 5 lim 4+-L- U + : an 2 8 J _ (4 + + 0) \ (1+0) 102. lim 0^0+ i - 2 cot 2 e 5 cot 2 8 - 1 cot 6 - 8 lim 0+ 5--J, 3|2fl7 (0-2) (5-0-0) 103. lim *° in * = lim xslnx , = lim xsm 2 x xU = lim 4^ • ^ x _, 2-2cosx x ^ 2(1- cos x) x ^ 2(2sin 2 (|)) x ^ [sin 2 (|) x lim X^ L Sln \2) S1 "l (1)(1)(1) = 1 104. lim l=f^ 0^0 " lim 0^0 2sin 2 (S) lim o L (I) P-^f - J] = dXl) (J) = J 105. lim ^ x^O x x^0 lim (^- • ^) = 1; let 6 = tan x => 9 -> as x -> =4> lim g(x) = lim ta " (tanx) > n v cos xx/ x > o x > x lim ^§-2 = 1. Therefore, to make g continuous at the origin, define g(0) = 1. 0^0 e Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 106. lim f(x) x — > lim tan (tan x) q sin (sin x) #105); let 6 = sin x =4> 6 lim x^O as x tan (tan x) tan x -> = continuous at the origin, define f(0) = 1 . sin (sin x) cos x ■ lim -^ys x — » sin ' sin x ) 1 • lim X — i lim Chapter 3 Practice Exercises 189 (using the result of n sin (sin x) ^3 = 1. Therefore, to make f ■ (a) S = 27rr 2 + 27rrh and h constant =>■ ^ = Airr | + 27rh (b) S = 27rr 2 + 27rrh and r constant => ^ = 27rr ^ at at (c) S = 2m 2 + 27rrh => f = 4tit g + 2tt (r f +h gj) = (d) S constant => f = => = (4?rr + 2?rh) g + 2?rr f (47TT + 27Th) g (47rr + 27rh)£+27rr£ (2r + h) dt dr dt dr _ -r dh dt 2r+h dt 108. S = TnVr 2 + h 2 (a) h constant = . dS ^ dt (b) r constant =4> g = ttt • (r f 2 +h | ) + Tr^r 2 + h 2 Vr 2 + h 2 dS dt Vr 2 +h 2 S _ 7rrh dh It ~~ Vr 2 + h 2 dt % + Trx/r 2 + h 2 g = [tt ^r 2 + h 2 + \/r 2 + h 2 J dt (c) In general, f = lir^/r 2 + h 2 + -T^jf—l g + -rf^ § W 6 ' dt [V y f2 + h 2 J dt y r 2 + h 2 dt 109. A = tit 2 => ^ = 27rr | ; so r = 10 and | = - f m/sec => ^ = (2tt)(10) (- f ) = -40 m 2 /sec 110. V = s 3 dV dt 3s 2 ds l = 3?f; sos = 2 0and dV dt 1200 cm 3 /min ds dt ~~ 3(20) 2 h* (1200) = 1 cm/min 111. SEl dt -1 ohm/sec, ^ = 0.5 ohm/sec; and ^ = ^ + ^ =* pf = Ff-sf' Also ' R Ri Ri = 75 ohms and R2 = 50 ohms fe(-D- (55? (0.5) -1 dR (30) 2 dt (75} 0.02 ohm/sec ,5625 - + - 75 T 50 500oJ = R = 30 ohms. Therefore, from the derivative equation, '- = (-900) ( 500Q ~ 5625 ^ — 9(625) 5625-5000 . 50(5625) J_ 50 112. <§ = 3 ohms/sec and <f = -2 ohms/sec; Z = ^R 2 + X 2 =>• f R I v dX f, d ; so that R ^R 2 + X2 10 ohms and X = 20 ohms dZ dt (10)(3)+(20)(-2) V / 10 2 + 2Q2 —0.45 ohm/sec. 113. Given ^7 = 10 m/sec and -# = 5 m/sec, let D be the distance from the origin =4> D 2 x 2 + y 2 2D dD 2x l + 2 y|^ D f = x l + y|' when ( x >y) = ( 3 ' - 4 )> D = ^ 2 + (-^f = 5 and (5)(10) + (12)(5) dD dt 110 5 22. Therefore, the particle is moving away from the origin at 22 m/sec (because the distance D is increasing). 1 14. Let D be the distance from the origin. We are given that dD dt 1 1 units/sec. Then D „sm 2D ^ ^ dt ,2 dx ZX d( -f JX dt x(2 + 3x) f ; x = 3 => D = ^3 2 + 3 3 and substitution in the derivative equation gives (2)(6)(1 1) = (3)(2 + 9) ^ % = 4 units/sec. dt 115. (a) From the diagram we have (b) V = i 7rr 2 h 1 nr I 1 I f l5 h)^ 47Th 3 75 dV dt |h. _ 47rh 2 dh dV — 25 dt ' SO dt -5 and h = 6 dh dt ma ft/min - 116. From the sketch in the text, s = rd ds dt dt dr dt Also r |=rf = (1.2) f. Th e refore,| 6 ft/sec andr = 1.2 ft 1 .2 is constant => g => f = 5 rad/sec Copyright (c) 2006 Pearson Education 190 Chapter 3 Differentiation 117. (a) From the sketch in the text, ^ -0.6 rad/sec and x = tan 0. Also x = tan 8 f = sec 2 6»f ;at point A, x = => 6 = => ^ = (sec 2 0) (-0.6) = -0.6. Therefore the speed of the light is 0.6 = | km/sec when it reaches point A. (b) ^^ • ^_ . §0sec = 18 revs/min v ' sec Z7r rad mm tt 118. From the figure, a = b => i = b — & ' r BC r ^(,2 — r 2 that r is constant. Differentiation gives, We are given 1 da r " dt ~~ b = 2r and db b 2 -r 2 -0.3r Then, da v ,:„---^(-a3r)-(2r)[-^ (2r) 2 - r 2 3r 2 (-0.3r)+^a PI _ (3r 2 )(-0.3r)+(4r 2 )(0.3r) _ 0.3r 3^ 3v/3 lOv^ m/sec. Since $ is positive, the distance OA is increasing when OB = 2r, and B is moving toward O at the rate of 0.3r m/sec. 1 19. (a) If f(x) = tan x and x = — \ , then f'(x) = sec 2 x, -1 andf i 1 V 4 1 * """ * V 4 f(x) is L(x) = 2 (x + r i) + (-1) = 2x 2. The linearization of t-2 2 (b) If f(x) = sec x and x = — | , then f '(x) = sec x tan x, f (- |) = v^andf (- |) =-\fl. The linearization of f(x) is L(x) = -a/2 (x + f ) + a/2 \/2(4-7r) 4 -tt/2 -it/4 . j. = -A&: + V2(4-lr|/4 120.f(x) l 1 + tan x f'(x) (1 +tanx) : . The linearization at x = is L(x) = f '(0)(x - 0) + f(0) = 1 - x. 121.f(x) = a/x~+T + sin x - 0.5 = (x + l) 1 / 2 + sin x - 0.5 =>• f'(x) = (i) (x + l)" 1 / 2 + cos x => L(x) = f'(0)(x - 0) + f(0) = 1.5(x - 0) + 0.5 =>■ L(x) = 1.5x + 0.5, the linearization of f(x). 122.f(x) = £ + a/1+x - 3.1 = 2(1 - x)- 1 + (1 + x) 1 / 2 - 3.1 => f'(x) = -2(1 - x)- 2 (-l) + \ (1 + x)" 1 / 2 - 2 | 1 (i-x) 2 "•" 2 yr + x L(x) = f'(0)(x - 0) + f(0) = 2.5x - 0.1, the linearization of f(x). 123. S = 7T rx/r 2 + h 2 , r ,2A-l/2 constant => dS = 7r r ■ ^(r 2 + h 2 ) ' 2h dh = ,,, , dh. Height changes from h to h + dh dS Trrho(dh) ^Th 2 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Chapter 3 Additional and Advanced Exercises 191 12r 2 124. (a) S = 6r 2 => dS = 12rdr. We want |dS| < (2%)S => |12rdr| < ±gl =>• |dr| < ^ . The measurement of the edge r must have an error less than 1%. (b) When V = r 3 , then dV = 3r 2 dr. The accuracy of the volume is (^) (100%) = (^f*) (100%) ;f) (dr)(100%) . 100. (100%) = 3% 125. C = 27rr => r = f , S = 47rr 2 = ^ , and V 4 —3 _ Ci 3 7rr _ 6?r2 It also follows that dr = ^ dC, dS = — dC and dV = A dC. Recall that C = 10 cm and dC = 0.4 cm. 2tH (a) dr=^ = ^cm (b) dS = f (0.4) = | cm =: (c) dV = (0.4) = | cm ;f ) (100%) = (^) (^) (100%) = (.04)(100%) = 4% k S . It J V 10 > (100%) = (I) (^) (100%) = 8% 'dV v V doo%) = (i)(^)(ioo%) 12% 126.Similar triangles yield ^ = T ^ h = 14 ft. The same triangles imply that ^±-2 = § 120 A t 120 \ / , 1 dh= -120a" 2 da da ±£J h = 120a- 1 + 6 if ) ( ± i) = ± 1 «rf ± .0444 ft = ± 0.53 inches. CHAPTER 3 ADDITIONAL AND ADVANCED EXERCISES 1. (a) sin 20 = 2 sin cos =4> ^ (sin 20) = ^ (2 sin cos 0) => 2 cos 20 = 2[(sin 0)(-sin 0) + (cos 0)(cos 0)] => cos 20 = cos 2 — sin 2 (b) cos 20 = cos 2 - sin 2 =>• j- g (cos 20) = ^ (cos 2 - sin 2 0) => -2 sin 20 = (2 cos 0)(-sin 0) - (2 sin 0)(cos 0) => sin 20 = cos sin + sin cos =4> sin 20 = 2 sin cos 2. The derivative of sin (x + a) = sin x cos a + cos x sin a with respect to x is cos (x + a) = cos x cos a — sin x sin a, which is also an identity. This principle does not apply to the equation x 2 — 2x — 8 = 0, since x 2 — 2x — 8 = is not an identity: it holds for 2 values of x (— 2 and 4), but not for all x. 3. (a) f(x) = cos x f'(x) -sin x =>• f"(x) = —cos x, and g(x) = a + bx + ex 2 =>■ g'(x) = b + 2cx = a = 1; f'(0) = g'(0) => -sin(0) = b => b = 0; f"(0) = g"(0) g"(x) = 2c; Therefore, g(x) = 1 — \ x 2 . '(x) = b cos x — c sin x; also, cos (a) = b cos (0) — c sin (0) also, f(0) = g(0) => cos (0) = a => -cos(0) = 2c => c = - \, (b) f(x) = sin (x + a) => f '(x) = cos (x + a), and g(x) = b sin x + c cos x = f(0) = g(0) => sin (a) = b sin (0) + c cos (0) => C = sin a; f '(0) = g'(0) => b = cos a. Therefore, g(x) = sin x cos a + cos x sin a. (c) When f(x) = cos x, f'"(x) = sin x and f ' 4 '(x) = cos x; when g(x) = 1 — I x 2 , g'"(x) = and g' 4 '(x) = Thus f '"(0) = = g'"(0) so the third derivatives agree at x = 0. However, the fourth derivatives do not agree since f ™(0) = 1 but g' 4 '(0) = 0. In case (b), when f(x) = sin (x + a) and g(x) = sin x cos a + cos x sin a, notice that f(x) = g(x) for all x, not just x = 0. Since this is an identity, we have f * n '(x) = g^°'( x ) f° r an Y x an d any positive integer n. = —sin x => y + y = -cos x + cos x = 0; y —sin x + sin x = 0; y = cos x =4> y' = —sin x a cos x + b sin x => y' = — a sin x + b cos x (a) y = sin x =4> y = cos x =>■ y =>• y" = —cos x => y" + y = - =>• y" = —a cos x — b sin x => y" + y = (—a cos x — b sin x) + (a cos x + b sin x) = (b) y = sin(2x) => y' = 2cos(2x) => y" = -4sin(2x) =>■ y" + 4y = -4 sin(2x) + 4 sin(2x) = 0. Similarly, y = cos (2x) and y = a cos (2x) + b sin (2x) satisfy the differential equation y' + 4y = 0. In general, y = cos (mx), y = sin (mx) and y = a cos (mx) + b sin (mx) satisfy the differential equation y" + m 2 y = 0. Copyright (c) 2006 Pearson Education 192 Chapter 3 Differentiation 5. If the circle (x — h) 2 + (y — k) 2 = a 2 and y = x 2 + 1 are tangent at (1, 2), then the slope of this tangent is = 2x| 2x 2 and the tangent line is y = 2x. The line containing (h, k) and (1, 2) is perpendicular to k-2 h- 1 h = 5 — 2k => the location of the center is (5 — 2k, k). Also, (x — h) 2 (y - k) 2 = a 2 =» x - h + (y - k)y' = 0^1 + (y') 2 + (y - k)y" y n _ i + (yT k-y At the point (1, 2) we know y' = 2 from the tangent line and that y" = 2 from the parabola. Since the second derivatives are equal at (1, 2) we obtain 2 1 + (2) 2 k-2 "^ iv 2 lies on the circle we have that a . Then h ~~ 2 ■ 2k the circle is (x + 4) 2 + (y I) 2 a 2 . Since (1,2) 6. The total revenue is the number of people times the price of the fare: r(x) = xp = x (3 — J| 40 ) < x < 60. The marginal revenue is j^ = (3 2x 3 dx ~~ V-' 40 40M* 40 V ' 1 "^" dx are on the bus the marginal revenue is zero and the fare is p(40) 40 ) 40 J dr dx 40 where )[(3- 40 ) 2x1 40 J 3 (3 ^)(1 tti) ■ Then * = =4> x = 40 (since x = 120 does not belong to the domain). When 40 people 40) $4.00. 7. (a) y = uv => ^ = ^ v + u ^ = (0.04u)v + u(0.05v) = 0.09uv = 0.09y =>- the rate of growth of the total production is 9% per year (b) If year dy f = -0.02u and g = 0.03v, then ^ = (-0.02u)v + (0.03v)u = O.Oluv = O.Oly, increasing at 1% per 8. When x 2 + y 2 = 225, then y' = - £ . The tangent line to the balloon at (12, -9) is y + 9 = f (x - 12) => y = | x — 25. The top of the gondola is 15 + 8 = 23 ft below the center of the balloon. The inter- section of y = —23 and y = | x — 25 is at the far right edge of the gondola -23 x-25 Thus the gondola is 2x = 3 ft wide. x 2 + y 2 = 225 (-12,-9) \ >, Suspension cables — \- Gondola y-(4/3)x-25 8ft L -Width NOT TO SCALE 9. Answers will vary. Here is one possibility. L 10. s(t) = 10 cos (t v(t) ds dt 10 cos 10 (a) s(0)- KAl (b) Left: -10, Right: 10 (c) Solving 10 cos (t+ |) = -10 Solving 10 cos (t+ I) = 10 = is farthest to the right. Thus, v (d) Solving 10 cos (t 10 sin t =^ cos (t - COS (t + 3tt> a(t) dv dt 3-a d£s dt- 1 10 cos (t — 1 => t = t when the particle is farthest to the left. 1 0, v 'llT\ 0, a 10, Ivi , but t > => t = 2-k - 10, and a ( 7 f) = -10. )\ = lOandaf?) = 0. — - 4 when the particle Copyright (c) 2006 Pearson Education Chapter 3 Additional and Advanced Exercises 193 11. (a) s(t) = 64t - 16t 2 =>• v(t) = f t = 64 - 32t = 32(2 - t). The maximum height is reached when v(t) = =>• t = 2 sec. The velocity when it leaves the hand is v(0) = 64 ft/sec. (b) s(t) = 64t - 2.6t 2 =>• v(t) = ^ = 64 - 5.2t. The maximum height is reached when v(t) = => t w 12.31 sec. The maximum height is about s(12.31) = 393.85 ft. 12. si = 3t 3 - 12t 2 + 18t + 5 and s 2 = -t 3 + 9t 2 - 12t =4> v : = 9t 2 - 24t + 18 and v 2 = -3t 2 + 18t - 12; v x = v 2 =^ 9t 2 - 24t + 18 = -3t 2 + 18t - 12 => 2t 2 - 7t + 5 = => (t - l)(2t - 5) = => t = 1 sec and t = 2.5 sec. 13. m(v 2 -v 2 )=k(x 2 -x 2 ) => m (2vf)=k(-2xf) => m| = k(-|) | => m £ = -kx (i) f . Then substituting $ = v =>• m ^ = — kx, as claimed. 14. (a) x = At 2 + Bt + C on [ti , t 2 ] 2At + B =>• v (^ j2 ) = 2A (^y^ 2 ) + B = A (ti + t 2 ) + B is the Ax At instantaneous velocity at the midpoint. The average velocity over the time interval is v a , _ (At| + Bt 2 + C)-(Atf + B tl + C) _ (t 2 - tl )[A(t 2 +ti)+B] = A U , j. \ , g fe — ti t 2 — ti ^ ' (b) On the graph of the parabola x = At 2 + Bt + C, the slope of the curve at the midpoint of the interval [ti , t 2 ] is the same as the average slope of the curve over the interval. 15. (a) To be continuous at x = 7r requires that lim_ sin x = lim (mx + b) => = van + b =>• m = — (b) if y' = { COS X X "-C 7T ' _ is differentiable at x = 7r, then lim_ cos x = m => m = — 1 and b = n. m, X > 7T X -> 7T 16. f(x) is continuous at because lim = f(0). f'(0) lim f(x) - f(0) lim iiL_0 lim 1 — COS X \ / 1 + COS X > X 2 / V 1 + COS X I lim (^ x^O V x ; , 1 + COS X ) x -° x-»"0 \ . Therefore f '(0) exists with value \ lim_ f(x) = f(2) f'(x) = | In order that f '(2) exist we must have a = 2a(2) — b =^> a = 4a — b 3a a = | and b 17. (a) For all a, b and for all x ^ 2, f is differentiable at x. Next, f differentiable at x = 2 =>• f continuous at x = 2 2a = 4a - 2b + 3 => 2a - 2b + 3 = 0. Also, f differentiable at x ^ 2 a, x < 2 2ax - b, x > 2 Then 2a — 2b + 3 = and 3a = b t t (b) For x < 2, the graph of f is a straight line having a slope of § and passing through the origin; for x > 2, the graph of f is a parabola. At x = 2, the value of the y-coordinate on the parabola is | which matches the y-coordinate of the point on the straight line at x = 2. In addition, the slope of the parabola at the match up point is | which is equal to the slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there. 18. (a) For any a, b and for any x/ — 1, g is differentiable at x. Next, g differentiable at x = — 1 => g continuous at x = — 1 =>- lim g(x) = g(— 1) =^ -a-l+2b = -a + b =>• b = 1. Also, g differentiable at x / - 1 X — » — 1 + I'M = { a, x < -1 3ax 2 + 1, x > -1 In order that g'(— 1) exist we must have a = 3a(— l) 2 + 1 =S> a=3a+l (b) For x < — 1, the graph of f is a straight line having a slope of — \ and a y-intercept of 1 . For x > — 1 , the graph of f is a parabola. At x = — 1 , the value of the y-coordinate on the parabola is | which matches the y-coordinate of the point on the straight line at x = — 1. In addition, the slope of the parabola at the match up point is — | which is equal to the slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there. 19. fodd f(-x) -f(x) d (f(-x)) = -f (-f(x)) =^ f'(-x)(-l) = -f'(x) => f'(-x) = f'(x) =>■ f is even. dx Copfiigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 194 Chapter 3 Differentiation 20. feven =► f(-x) = f(x) =► A (f(- x )) = A (f(x)) f'(-x)(-l) = f'(x) => f (-x) = -f (x) f is odd. 21. Leth(x) = (fg)(x) = f(x)g(x) => h'(x) = lim A > AQ h(x) - h(x ) _ j^ m x - x x — > x f(x)g(x)-f(x )g(x ) X-XQ lim x^ x -x^xo [fwl^f 21 ]] +x 1 iPl h [ ^ il f(x) g(x) - f(x) g(x ) + f(x) g(x ) - f(x ) g(x ) f(xo) x lim o «[=*U + g(x )f (xo) = ■ x lim Q x-*o J x —> x L DX "' |_ x ~ x o I ^5^1 + g(x ) f (xo) = g(x ) f '(Xq), if g is continuous at Xq. Therefore (fg)(x) is differentiable at Xq if f(xg) = 0, and (fg)' (xq) = g(xg) f '(xq). 22. From Exercise 21 we have that fg is differentiable at if f is differentiable at 0, f(0) = and g is continuous atO. (a) If f(x) = sin x and g(x) = |x| , then |x| sin x is differentiable because f'(0) = cos (0) = 1, f(0) = sin (0) = and g(x) = |x| is continuous at x = 0. (b) If f(x) = sin x and g(x) = x 2 / 3 , then x 2 / 3 sin x is differentiable because f'(0) = cos (0) = 1, f(0) = sin (0) = and g(x) = x 2 / 3 is continuous at x = 0. (c) If f(x) = 1 — cos x and g(x) / x, then \/x(l — cos x) is differentiable because f'(0) = sin (0) = 0, f(0) = 1 cos (0) = and g(x) = x 1//3 is continuous at x = 0. (d) If f(x) = x and g(x) = x sin Q) , then x 2 sin Q) is differentiable because f'(0) = 1, f(0) = and (so g is continuous at x = 0). lim x sin (-) x^0 Vx; lim x^0 lim ^ t — ► oo ' 23. If f(x) = x and g(x) = x sin (-) , then x 2 sin (-) is differentiable at x = because f'(0) = 1, f(0) = and lim x sin (-^ x^0 Vx/ lim ffi lim t — » 00 (so g is continuous at x = 0). In fact, from Exercise 21, h'(0) = g(0)f'(0) = 0. However, forx ^ 0, h'(x) = [x 2 cos (1)] (- ^) + 2x sin Q) . But lim h'(x) = lim [—cos (-) + 2x sin (i)l does not exist because cos (-) has no limit as x — > 0. Therefore, x^O x^O L Vx7 Vx;J VxV the derivative is not continuous at x = because it has no limit there. 24. From the given conditions we have f(x + h) = f(x) f(h), f(h) 1 = hg(h) and lim g(h) = 1. Therefore, h — > f'(x) = lim f(x+h) - f(x) lim f(x)f(h)-f(x) lim f(x) [Sflpi h -► L h f(x) lim g(h) Ln — > u f(x) • 1 = f(x) f'(x) = f(x) and f'(x)exists at every value of x. dy 25. Step 1: The formula holds for n = 2 (a single product) since y = U1U2 - , Step 2: Assume the formula holds for n = k: dv dui 1 dn~ y = uiu 2 ---u k => ai = dF u 2 u 3 ---u k + Ify = uiu 2 ---u k u k+ i = (u 1 u 2 ---u k )u k+1 ,then ^ = ; iv ' u k _ £ U 2 + Ui U1II2— Uti £ (^u 2 u 3 ---u k + Ui ^ u 3 ---u k + ••• ^U 2 U 3 ---U k+1 +Ul ^U 3 ---U k+1 + Hi], ^ UiU 2 ---U k _, -j±) lit dx din dx , UlU 2 ---U k ^±l u lU2 -u k ^ UiU 2 ---U k du. + U!U 2 ---U! du,. Thus the original formula holds for n = (k+1) whenever it holds for n = k. 26. Recall „ ™' , ■ Then(™ k! (m — k)! VI. l!(m-l)! m and m!(k+l) + m!(m-k) m!(m+ 1) (m+1)! (k+l)!(m-k)! (k+l)!(m-k)! (k+ 1)! ((m + 1) - (k+ 1))! Leibniz's rule by mathematical induction. Stepl: Ifn=l,then^ Vk+W k!(m-k)! [k+i ) ■ Now, we prove (k+l)!(m-k-l)! U 3x + v 3x ■ Assume that the statement is true for n = k, that is: d'-(iiv) dx 1 d k u .. I u d k ] u dv i /k\ d k 2 u d 2 v i l V ~t~ K A„k~l Av ' \ 1 I ^»k-2 A^2 T Step 2: Ifn = k+1, then dx k dx k_1 dx d k+ '(uv) ,2) dx k - 2 dx 2 d_ dx + k \ du d k ~'v k - l) dv dx k -' d*v dx k I d k (uv) \ __ r d k +'u , d"u dv] , |V d k u dv , ^ d k ~'u d 2 v ] \ dx k ) ~ Ldx k +! v "t" dx k dxJ + [ K dx k dx + K dx k -i dx 2 J Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle Chapter 3 Additional and Advanced Exercises 195 k\ d k -'u d 2 v , .2) dx k -i dx 2 " t " du ft 1 d k+1 u ] dx dx k "r" u dx k +ij 'k\ d k ~ 2 u d 3 v ^2/ dx k - 2 dx 3 ' k \ ft d k ~'v ,k-lJ dx 2 dx k -i \ k— 1 / dx dx 1 du d u .. + [ + [( k *i) + (*)]££ + d^v + (k+i) du dv dx k dx [(l) + (!)]DS 'k+l\ du dS; \ k / dx dx k du d^v dx dx k d k+1 v d ki 'v dx 1 * 1 ^v + (k+l)0^ 'k+l\ d M u d 2 v dx 1 - dx-' + . U dx k +! • Therefore the formula (c) holds for n = (k + 1) whenever it holds for n = k. 27. (a) T 2 = ^ ^ L = ^ L = Lligg^ztt^j ^ L ^ agl5fi ft (b) T 2 = ^ ^ T = ^Lx/l; dT = ^L . 1 dL = -^dL; dT = ,, noir , J 1 ^^, ,, (0.01 ft) « 0.00613 sec. v ' g x/g x/g 2x/L x/Lg x/(0.8156ft)(32.2 ft/sec 2 ) v ' (c) Since there are 86,400 sec in a day, we have (0.00613 sec)(86,400 sec/day) « 529.6 sec/day, or 8.83 min/day; the clock will lose about 8.83 min/day. = s 3 2k dt so - 3s 2ds -k(6s 2 ) dt 2k. If So = the initial length of the cube's side, then Si = &o — 2k (vo) 1/3 (Ml 3 Si. Let t = the time it will take the ice cube to melt. Now, t = S 11 hr. so S0-S1 -(boY Copyright (c) 2006 Pearson Education 196 Chapter 3 Differentiation NOTES: Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle CHAPTER 4 APPLICATIONS OF DERIVATIVES 4.1 EXTREME VALUES OF FUNCTIONS 1. An absolute minimum at x = C2, an absolute maximum at x = b. Theorem 1 guarantees the existence of such extreme values because h is continuous on [a, b]. 2. An absolute minimum at x = b, an absolute maximum at x = c. Theorem 1 guarantees the existence of such extreme values because f is continuous on [a, b]. 3. No absolute minimum. An absolute maximum at x = c. Since the function's domain is an open interval, the function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values. 4. No absolute extrema. The function is neither continuous nor defined on a closed interval, so it need not fulfill the conclusions of Theorem 1. 5. An absolute minimum at x = a and an absolute maximum at x = c. Note that y = g(x) is not continuous but still has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur. 6. Absolute minimum at x = c and an absolute maximum at x = a. Note that y = g(x) is not continuous but still has absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur. 7. Local minimum at ( — 1, 0), local maximum at (1, 0) 8. Minima at (-2, 0) and (2, 0), maximum at (0, 2) 9. Maximum at (0, 5). Note that there is no minimum since the endpoint (2, 0) is excluded from the graph. 10. Local maximum at (—3, 0), local minimum at (2, 0), maximum at (1, 2), minimum at (0, —1) 11. Graph (c), since this the only graph that has positive slope at c. 12. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c. 13. Graph (d), since this is the only graph representing a funtion that is differentiable at b but not at a. 14. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b. Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 198 Chapter 4 Applications of Derivatives 15. f(x) = | x - 5 => f (x) f(-2) = - ¥, f(3) = -3 => no critical points; the absolute maximum 19 is —3 at x = 3 and the absolute minimum is — y at 16. f(x) = —x — 4 =>- f (x) = — 1 => no critical points; f(— 4) = 0, f(l) = — 5 =>• the absolute maximum is at x = —4 and the absolute minimum is —5 at x = 1 (1,-5) 17. f(x) = x 2 — 1 =>• f'(x) = 2x =4* a critical point at x = 0; f(-l) = 0, f(0) = -1, f(2) = 3 => the absolute maximum is 3 at x = 2 and the absolute minimum is — 1 atx = 18. f(x) = 4 - x 2 =4> f (x) = -2x => a critical point at x = 0; f(-3) = -5, f(0) = 4, f(l) = 3 => the absolute maximum is 4 at x = and the absolute minimum is —5 at x = — 3 /(x) = 4-x 19. F(x) F'(x) = 2x" , however x = is not a critical point since is not in the domain; F(0.5) = —4, F(2) = —0.25 => the absolute maximum is —0.25 at x = 2 and the absolute minimum is —4 at x = 0.5 1 v 1 . (2,-0.25) -1 Abs max -2 jy = 4;,0.5<i<2 X -3 -4 - • (0.5, -4) Abs nin Copyright (c) 2006 Pearson Education Section 4. 1 Extreme Values of Functions 199 20. F(x) F'(x) , however x = is not a critical point since is not in the domain; F(— 2) = | , F(— 1) = 1 =£- the absolute maximum is 1 at x = — 1 and the absolute minimum is k at x = — 2 (-1,1) (-2,1/2) F(x) = — -15 -1 -0.5 y i 21. h(x) = \/x = x 1 / 3 =>■ h'(x) I ,,-2/3 3 A a critical point at x = 0; h(-l) = -1, h(0) = 0, h(8) = 2 =>• the absolute maximum is 2 at x = 8 and the absolute minimum is — 1 at x = - 1 2 _ -1<.t<8 „ (8. 2) 1 V^"""^ max M (-1,-D Abs min 12 3 4 5 6 7 8 22. h(x) = -3x 2 / 3 => h'(x) = -2x -1 / 3 => a critical point at x = 0; h(-l) = -3, h(0) = 0, h(l) = -3 => the absolute maximum is at x = and the absolute minimum is —3 at x = 1 and at x = — 1 (-1,-3) h(x) = -3* 2/3 (1,-3) 23. g(x) = \/4-x 2 = (4 - x 2 ) 1/2 => g'(x) = i (4 - x 2 )" 1/2 (-2x) v/4-x 2 =>- critical points at x = —2 and x = 0, but not at x = 2 because 2 is not in the domain; g(— 2) = 0, g(0) = 2, g(l) = V 3 =>■ the absolute maximum is 2 at x = and the absolute minimum is at x = —2 t * y = ^4-^ (0, 2) Abs max -2<.*<1^ 1 - (-2,0) - 1 Abs min 24. g(x) = -A/5-x 2 = (5 x 2 ) 1/2 (5 - x 2 )~ 1/2 (-2x) => critical points at x = — y 5 and x = 0, but not at x = y 5 because y 5 is not in the domain; f (- V^) = 0, f(0) = -\/5 =>■ the absolute maximum is at x = — y 5 and the absolute minimum is — y 5 at x = -2.5 -2 -1.5 -1 .0.5 (-/5\0) «0O = -V5^7 (0,-v^) 25. f(0) = sin i f (0) = cos (9 =>■ = f is a critical point, but = -rp is not a critical point because -f is not interior to the domain; f ( -i,f(!) = i,f($) =4> the absolute maximum is 1 at 6* = | and the absolute minimum is — 1 at = =2 (-ir/2,-1) Abs min (tt/2, 1) Abs max -J— ■nil 5w!6 y=sm8,-ir/2<0< 5irl6 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 200 Chapter 4 Applications of Derivatives 26. f(9) = tan 9 => f'(9) = sec 2 => f has no critical points in (^p, ^) . The extreme values therefore occur at the endpoints: f (=^) = — y3andf (|) = 1 =^ the absolute maximum is 1 at 6 = | and the absolute minimum is — y 3 at — 7T 3 ir/4,1) -it/3, -/T) 27. g(x) = esc x =£- g'(x) = —(esc x)(cot x) => a critical point at x 73-SV2, i.g(¥) the absolute maximum is -7- at x = f and x = 7? , and the absolute minimum is 1 at x = 5 Abs max Abs max (tt/3, 2/Vl) (2tt/3, 2/VI) J-CSC.I (77/2,1) 77/3<^<277/3 ^ bs min 77/3 77/2 277/3 28. g(x) = sec x => g'(x) = (sec x)(tan x) =>■ a critical point at the absolute x = 0;g(-f)=2,g(0)=l,g(f) = ^ maximum is 2 at x atx = I and the absolute minimum is 1 (-tt/3,2 (Tt/6,2/vf) g (jr) s sec* -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 2\l/2 29. f(t) = 2 — |t| = 2 — a/^" = 2 — (t 2 ) ^f(t)=-i(t 2 r 1/2 (2t) = --^ =S> a critical point at t = 0; f(— 1) = 1, f(0) = 2, f(3) = 1 =>• the absolute maximum is 2 at t = and the absolute minimum is — 1 at t = 3 (0, 2) Abs \ max » 1 \ y = 2-l/l \ -1<(<3 -1 1 ZV 3 -1 AbsN^ nun (3] _i) 30. f(t) = |t - 5 1 = ^(T^ = (ft - 5ff 2 => f ft) i((t-5) 2 )" 1/2 (2(t-5)) t-5 |t-5| 2 VV "> I \^ ->!> ^(1 -5)2 =4> a critical point at t = 5; f(4) = 1, f(5) = 0, f(7) = 2 =>■ the absolute maximum is 2 at t = 7 and the absolute minimum is at t = 5 f(t) = \t-5\ 31. f(x) = x 4 / 3 => f (x) = I x 1 / 3 =>• a critical point at x = 0; f(- 1) = 1, f(0) = 0, f(8) = 16 =>■ the absolute maximum is 16 at x = 8 and the absolute minimum is at x = 32. f(x) = x 5 / 3 => f'(x)=fx 2 / 3 =>• a critical point at x = 0; f(- 1) = -1, f(0) = 0, f(8) = 32 =>■ the absolute maximum is 32 at x = 8 and the absolute minimum is — 1 at x = — 1 33. g(<9) = 6> 3 / 5 => g'(6)= \9- 2 l 5 -, a critical point at = 0; g(— 32) maximum is 1 at 9 = 1 and the absolute minimum is — 8 at 9 = —32 -8, g(0) = 0, g(l) = 1 => the absolute Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle Section 4. 1 Extreme Values of Functions 201 34. h(0) = 30 2 / 3 => h'(0) = IQ- 1 ^ => a critical point at = 0; h(-27) = 27, h(0) = 0, h(8) = 12 => the absolute maximum is 27 at = —27 and the absolute minimum is at 8 = 35. Minimum value is 1 at x = 2. -2 -2 [-2,6] by [-2,4] 36. To find the exact values, note that y' = 3x 2 — 2, which is zero when x = ± Local maximum at -,/§,4 4y/C 9 (-0.816, 5.089); local minimum at (\/i> 4 _ ^) ~ (°- 816 ' 2 - 911 ) 6,6] by [-2,7] 37. To find the exact values, note that that y' = 3x 2 + 2x — 8 = (3x — 4)(x + 2), which is zero when x = — 2 or x = |. Local maximum at (—2, 17); local minimum at (I, .in 27 1 2 4 6 6,6] by [-5,20] 38. Note that y' = 3x 2 - 6x + 3 = 3(x - l) 2 , which is zero at x = 1. The graph shows that the function assumes lower values to the left and higher values to the right of this point, so the function has no local or global extreme values. 6,6] by [-4,4] 39. Minimum value is when x - 1 or x = 1 . Cop # (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 202 Chapter 4 Applications of Derivatives 40. The minimum value is 1 at x = 0. -l l [-1.5,1.5] by [-0.5,3] 41. The actual graph of the function has asymptotes at x = ± 1, so there are no extrema near these values. (This is an example of grapher failure.) There is a local minimum at (0,1). [-4.7,4.7] by [-3.1,3.1 42. Maximum value is 2 at x = 1; minimum value is at x = —1 and x = 3. [-4.7,4.7] by [-3.1,3.1] 43. Maximum value is \ at x = 1; minimum value is — | as x = — 1. -5,5] by [-0.7,0.7] 44. Maximum value is \ at x = 0; minimum value is — \ as x = —2. -5,5] by [-0.8,0.6] Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 4. 1 Extreme Values of Functions 203 45. y' = x 2 / 3 (l) + fx-^x + 2 ) = |+i cnt. pt. derivative undefined extremum local max local min value iflO 1 / 3 = 1.034 -3 x [-4,4] by [-3,3] 46. 8x 2 -8 y' = x 2 / 3 (2x) + |x- 1 / 3 (x 2 - 4) = s -f0 crit. pt. derivative extremum value x= -1 x = x= 1 undefined minimum local max minimum -3 3 y \ 2 \ 1 / 4 ^2 i 2 4 \ -i/\ j v^ 3 1 vy -4,4] by [-3,3] 47. y' = x -l ( - 2x) + (1)^4 -x2 ^2^4-x 2 -x 2 + (4-x 2 ) 4-2x 2 V4 — x 2 V4 — x 2 crit. pt. derivative extremum value x= -2 undefined local max x = -y2 minimum -2 x= s/2 maximum 2 x = 2 undefined local min h 3 - 2 - .. 1 •^ \ "1 ' ~7 ' ' * -2 -1 / 1 2 ^ -^ -2 - -3 - [-2.35,2.35] by [-3.5,3.5] -x 2 + (4x)(3-x) _ _5x 2 + 12x 2%/3-x " 2^/3 -x crit. pt. 12 5 3 derivative undefined extremum minimum local max minimum value 144 15 l/2 125 LO 4.462 [-4.7,4.7] by [-1,5] 49. y' -2, x<l 1, x> 1 crit. pt. derivative extremum value x= 1 undefined minimum 2 -2 2 4 -4.7,4.7] by [0,6.2] Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 204 Chapter 4 Applications of Derivatives 50 y'=( - 1 ' X< ° y 1 2 - 2x, x > crit. pt. derivative extremum value x = x= 1 undefined local min local max 3 4 [-4,4] by [-1,6] 51. y' -2x - 2, -2x + 6, x< 1 x> 1 crit. pt. derivative extremum value x= -1 x= 1 x = 3 undefined maximum local min maximum 5 1 5 -4,6] by [-2,6] f -I x 2 - I X + 11 52. We begin by determining whether f'(x) is defined at x = 1, where f(x) = < 4 2 4 [x 6 - 6x 2 + 8x, x< 1 x> 1 Clearly, f'(x) = -H- \ if x < 1, and lim f'(l + h) = -1. Also, f'(x) = 3x 2 - 12x + 8ifx> 1, and lim f ' ( 1 + h) = — 1 . Since f is continuous at x = 1 , we have that f ' ( 1 ) h->0+ -l.Thus, f'(x) 3x 2 - 12x + I x < 1 x> 1 Note that -\n- \ = when x = -1, and 3x 2 - 12x + 1 when x 12±yi2 2 -4(3)(8) _ 12±y / 48 But 2 2y/3 0.845 < 1, so the critical points occur at x = —1 and x = 2 2(3) 1-sfl „ 2± 2^3 3.155. crit. pt. derivative extremum value x = -l x* 3.155 local max local min 4 » -3.079 -4,6] by [-5,5] 53. (a) No, since f'(x) = |(x - 2) 1/3 , which is undefined at x = 2. (b) The derivative is defined and nonzero for all x/2. Also, f(2) = and f(x) > for all x^2. (c) No, f(x) need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed interval of the form [a, b] would have both a maximum value and minimum value on the interval. (d) The answers are the same as (a) and (b) with 2 replaced by a. 54. Note that f( »> = { -x 3 + 9x, x< -3or0 <x<3 ^ c rl , . f -3x 3 + 9, x<-3or0<x<3 , . Therefore, fix) = < ., x 3 -9x, -3<x<0orx>3 w \ 3x 3 - 9, -3<x<0orx>3 (a) No, since the left- and right-hand derivatives at x = 0, are —9 and 9, respectively. (b) No, since the left- and right-hand derivatives at x = 3, are —18 and 18, respectively. Copyright (c) 2006 Pearson Education Section 4. 1 Extreme Values of Functions 205 (c) No, since the left- and right-hand derivatives at x = —3, are 18 and —18, respectively. (d) The critical points occur when f (x) = (at x = ± \J 3) and when f (x) is undefined (at x = and x = ±3). The minimum value is at x = —3, at x = 0, and at x = 3; local maxima occur at ( — v 3, 6\/3) and ( \/3, 6v3). 55. DF -H^H •R x i 9-x (a) The construction cost is C(x) a graph of C(x). O.Z^/AS 0.2(9 — x) million dollars, where < x < 9 miles. The following is ^^ 2.95 J2 M 1 2.9 2.85 £ o 2.8 2.75 2.7 2.65 8 9 Solving C'(x) 0.3x V16 + X2 0.2 = gives x = ± 8\A 4 5 6 x (miles) w ±3.58 miles, but only x = 3.58 miles is a critical point is the specified domain. Evaluating the costs at the critical and endpoints gives C(0) = $3 million, CI -^- J « $2,694 million, and C(9) « $2,955 million. Therefore, to minimize the cost of construction, the pipeline should be placed from the docking facility to point B, 3.58 miles along the shore from point A, and then along the shore from B to the refinery, (b) If the per mile cost of underwater construction is p, then C(x) = p\/16 + x 2 + 0.2(9 — x) and C(x) = , 0,3 * ^ — 0.2 = gives x c = , ° ,s , which minimizes the construction cost provided x c < 9. The value v ; \/l6 + x 2 & c yV-0.04' F c — of p that gives x c = 9 miles is 0.218864. Consequently, if the underwater construction costs $218,864 per mile or less, then running the pipeline along a straight line directly from the docking facility to the refinery will minimize the cost of construction. In theory, p would have to be infinite to justify running the pipe directly from the docking facility to point A (i.e., for x c to be zero). For all values of p > 0.218864 there is always an x c S (0, 9) that will give a minimum value for C. This is proved by looking at C"(x c ) = ^—^ which is always positive for p > 0. 56. There are two options to consider. The first is to build a new road straight from Village A to Village B. The second is to build a new highway segment from Village A to the Old Road, reconstruct a segment of Old Road, and build a new highway segment from Old Road to Village B, as shown in the figure. The cost of the first option is Ci = 0.5(150) million dollars = 75 million dollars. Copyright (c) 2006 Pearson Education 206 Chapter 4 Applications of Derivatives New Construction Upgrade 150-2* Old Road The construction cost for the second option is C2(x) = 0.5 ( 2\/2500 + x 2 1 + 0.3(150 — 2x) million dollars for < x < 75 miles. The following is a graph of C2(x). Solving C' 2 (x) V2500 - 0.6 S 10 15 20 25 30 35 40 45 SO 55 80 65 70 75 x(milas) give x = ± 37.5 miles, but only x = 37.5 miles is in the specified domain. In summary, Ci = $75 million, C 2 (0) = $95 million, C 2 (37.5) = $85 million, and C 2 (75) a new road straight from village A to village B is the least expensive option. ).139 million. Consequently, 57. 10-Jt D The length of pipeline is L(x) = y 4 + x 2 + y 25 + (10 — x) for < x < 10. The following is a graph of L(x) Setting the derivative of L(x) equal to zero gives L'(x) (10 -x) 0. Note that \/4 + x 2 cos 9a and \A + x 2 ^25 + (10-x) 2 — — 12^2 = cos 6> B , therefore, L'(x) = when cos A = cos 9 B , or 6 A = 8 B and AACP is similar to ABDP. Use \/25 + (10-x) 2 simple proportions to determine x as follows: I = iSjp =>• x = y w 2.857 miles along the coast from town A to town B. If the two towns were on opposite sides of the river, the obvious solution would be to place the pump station on a straight line (the shortest distance) between two towns, again forcing #a = #b ■ The shortest length of pipe is the same regardless of whether the towns are on thee same or opposite sides of the river. Copyright (c) 2006 Pearson Education Section 4. 1 Extreme Values of Functions 207 58. 50 ft 30 ft (a) The length of guy wire is L(x) = ^900 + x 2 + y/2500 + (150 - x) 2 for < x < 150. The following is a graph of L(x). Setting L'(x) equal to zero gives L'(x) V^OO + x2 ^2500 + (150 -x) 2 0. Note that V900 + X 2 cos 8a and — — - = cos #b- Therefore, L'(x) = when cos 9 a = cos 9b, or 9 a = 9b and A ACE is similar to AABD. \/2500 + (150-x) Use simple proportions to determine x: ^ 150 -x 225 56.25 feet. 50 ^ 4 (b) If the heights of the towers are he and he, and the horizontal distance between them is s, then L(x) = Jh 2 c + x 2 + \/h 2 + (s - x) 2 and L'(x) (s-x) -. However, cos 9c and i ( s ~ x > = cos #b- Therefore, L'(x) = when cos 9r = cos 9b, or 9q = 9b and A ACE is similar to AABD. \/hB + (s-x) 2 S — X he *=(^) f Simple proportions can again be used to determine the optimum x: £■ 59. (a) V(x) = 160x - 52x 2 + 4x 3 V'(x) = 160 - 104x + 12x 2 = 4(x - 2)(3x - 20) The only critical point in the interval (0, 5) is at x = 2. The maximum value of V(x) is 144 at x = 2. (b) The largest possible volume of the box is 144 cubic units, and it occurs when x = 2 units. 60. (a) P'(x) = 2 - 200x~ 2 The only critical point in the interval (0, oo) is at x = 10. The minimum value of P(x) is 40 at x = 10. (b) The smallest possible perimeter of the rectangel is 40 units and it occurs at x = 10 units which makes the rectangle a 10 by 10 square. 61. Let x represent the length of the base and y 25 — x 2 the height of the triangle. The area of the triangle is represented by A(x) = |y25 — x 2 where < x < 5. Consequently, solving A'(x) 25 - 2x 2 2^25 -x2 Since A(0) = A(5) = 0, A(x) is maximized at x . The largest possible area is A ( 5 \ _ 25 -f cm' Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 208 Chapter 4 Applications of Derivatives 62. (a) From the diagram the perimeter P = 2x 4- 27rr = 400 => x = 200 — 7rr. The area A is 2rx => A(r) = 400r - 2?rr 2 where < r < ^. (b) A'(r) = 400 - 47rr so the only critical point is r = ™. Since A(r) = if r = and x = 200 - irr = 0, the values r = — w 31.83 m and x = 100 m maximize the 7T area over the interval < r < — . C3 63. s = -|gt 2 + v t + s => f t = -gt + v = 0=> t = ^.Nows(t) = s <3>t(-§ + v ) = <^> t = Oort Thuss(^)=-ig(^) 2 + vo(f)+s 2vo g 2\'i) 64. f t = -2sin t + 2cos t, solving | =>■ tan t = 1 =4> t never negative) =4> the peak current is 2y 2 amps. Y + So > so is the maximum height over the interval < t < | + nir where n is a nonnegative integer (in this exercise t is 65. Yes, since f(x) = |x| = yx 2 = (x 2 ) 2\l/2 f'(x) 1 ,2x-l/2 (2x) 2 V" I v~v (x2) i/2 | x is not required that f be zero at a local extreme point since f ' may be undefined there. is not defined at x = 0. Thus it 66. If f(c) is a local maximum value of f, then f(x) < f(c) for all x in some open interval (a, b) containing c. Since f is even, f(— x) = f(x) < f(c) = f(— c) for all — x in the open interval (— b, —a) containing — c. That is, f assumes a local maximum at the point — c. This is also clear from the graph of f because the graph of an even function is symmetric about the y-axis. 67. If g(c) is a local minimum value of g, then g(x) > g(c) for all x in some open interval (a, b) containing c. Since g is odd, g(— x) = — g(x) < — g(c) = g(— c) for all — x in the open interval (— b, —a) containing — c. That is, g assumes a local maximum at the point — c. This is also clear from the graph of g because the graph of an odd function is symmetric about the origin. 68. If there are no boundary points or critical points the function will have no extreme values in its domain. Such functions do indeed exist, for example f(x) = xfor— oo<x<oo. (Any other linear function f(x) = mx + b with m/0 will do as well.) 69. (a) f '(x) = 3ax 2 + 2bx + c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f. The function f(x) 3x has two critical points atx = —1 and x = 1. The function f(x) = x 3 — 1 has one critical point atx = 0. The function f(x) = x 3 + x has no critical points. (b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the cubic function has no extreme values.) Copyright (c) 2006 Pearson Education Section 4. 1 Extreme Values of Functions 209 70. (a) -0.1,0.6] by [-1.5,1.5] f(0) = is not a local extreme value because in any open interval containing x = 0, there are infinitely many points where f(x) = 1 and where f(x) = — 1. (b) One possible answer, on the interval [0, 1]: f(x) = j(l-*)cos^, 0<x<l This function has no local extreme value at x = 1. Note that it is continuous on [0, 1]. 7 1 . Maximum value is 11 at x = 5; minimum value is 5 on the interval [—3, 2]; local maximum at ( — 5, 9) 72. Maximum value is 4 on the interval [5, 7]; minimum value is —4 on the interval [—2, 1]. y 12" 10 / \ 8 ' / \ 6 / 4 2 -6,6] by [0,12] [-3,8] by [-5,5] 73. Maximum value is 5 on the interval [3, oo); minimum value is —5 on the interval (— oo, —2]. 6,6] by [-6,6] Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 210 Chapter 4 Applications of Derivatives 74. Minimum value is 4 on the interval [— 1, 3] \ / \ 8 / \ 6 " / / 2 [-6,6] by [0,9] 75-80. Example CAS commands: Maple : with(student): f := x -> x A 4 - 8*x A 2 + 4*x + 2; domain := x=-20/25.. 64/25; plot( f(x), domain, color=black, title="Section 4.1 #75(a)" ); Df := D(f); plot( Df(x), domain, color=black, title="Section 4.1 # 75(b)" ) StatPt := fsolve( Df(x)=0, domain ) SingPt := NULL; EndPt := op(rhs(domain)); Pts :=evalf([EndPt,StatPt,SingPt]); Values := [seq( f(x), x=Pts )]; Maximum value is 2.7608 and occurs at x=2.56 (right endpoint). Minimum value J is -6.2680 and occurs at x=l. 86081 (singular point). Mathematica : (functions may vary) (see section 2.5 re. RealsOnly ): «Miscellaneous "RealOnly" Clear[f,x] a=-l ; b= 10/3; f[x_] =2 + 2x - 3 x 2/3 f[x] Plot[{f[x],f[x]},{x,a,b}] NSolve[f [x]==0, x] {f[a],f[0],f[x]/.%,f[b]//N In more complicated expressions, NSolve may not yield results. In this case, an approximate solution (say 1.1 here) is observed from the graph and the following command is used: FindRoot[f[x]==0,{x, 1.1}] 4.2 THE MEAN VALUE THEOREM 1. When f(x) = x 2 + 2x - 1 for < x < 1, then ^'ig ' = f'(c) =S> 3 = 2c + 2 => c: 2. When f(x) = x 2 / 3 for < x < 1, then f H±dp. = f'( c ) =^> 1 = (§) c" 1 / 3 => c = ±. 3. When f(x) = x + 1 for \ < x < 2, then f(2 .^ /2) = f (c) ^0=l-^^>c=l. 4. When f(x) = yjx - 1 for 1 < x < 3, then ^=fl = f( c ) => & - --'-- - 2Vc-l 5. Does not; f(x) is not differentiable at x = in (—1, 8). Gopyifyt (c) 1 Pearson Etation, Inc., publishing as Pearson Addison-Wesle Section 4.2 The Mean Value Theorem 21 1 6. Does; f(x) is continuous for every point of [0, 1] and differentiable for every point in (0, 1). 7. Does; f(x) is continuous for every point of [0, 1] and differentiable for every point in (0, 1). 8. Does not; f(x) is not continuous at x = because lim_ f(x) = 1 / = f(0). x — > 9. Since f(x) is not continuous on < x < 1, Rolle's Theorem does not apply: lim_ f(x) = lim_ x = 1 X — > 1 X —> 1 ¥= o = f(i). 10. Since f(x) must be continuous at x = and x = 1 we have lim f(x) = a = f(0) => a = 3 and x^0+ lim f(x) = lim f(x) =>• — l+3 + a = m + b =>• 5 = m + b. Since f(x) must also be differentiable at x -> 1 x^ 1+ x = 1 we have lim f'(x) = lim f'(x) => — 2x + 3| , = ml , x-> I" V ' x^ 1+ lx=1 U=1 1 = m. Therefore, a = 3, m = 1 and b = 4. 11. (a) i n iii iv — • -2 — • •— 2 *x -5 -4 -3 — •— -1 • — • 2 — • • >JC (b) Let Xi and r 2 be zeros of the polynomial P(x) = x" + a^x' 1 " 1 + . . . + a x x + a , then P(rj) = P(r 2 ) = 0. Since polynomials are everywhere continuous and differentiable, by Rolle's Theorem P'(r) = for some r between ri and r 2 , where P'(x) = nx"" 1 + (n — 1) a n .|X"~ 2 + . . . + ai. 12. With f both differentiable and continuous on [a, b] and f(ri) = f(r 2 ) = ffo) = where ri, r 2 and r3 are in [a, b], then by Rolle's Theorem there exists a ci between ri and r 2 such that f'(ci) = and a c 2 between r 2 and r3 such that f'(c 2 ) = 0. Since f ' is both differentiable and continuous on [a, b], Rolle's Theorem again applies and we have a C3 between Ci and c 2 such that f ''(C3) = 0. To generalize, if f has n+ 1 zeros in [a, b] and f w is continuous on [a, b], then f W has at least one zero between a and b. 13. Since f " exists throughout [a, b] the derivative function f ' is continuous there. If f has more than one zero in [a, b], say f '(ri) = f'(r 2 ) = for ri ^ r 2 , then by Rolle's Theorem there is a c between ri and r 2 such that f "(c) = 0, contrary to f" > throughout [a, b]. Therefore f ' has at most one zero in [a, b]. The same argument holds if f" < throughout [a, b]. 14. If f(x) is a cubic polynomial with four or more zeros, then by Rolle's Theorem f '(x) has three or more zeros, f "(x) has 2 or more zeros and f'"(x) has at least one zero. This is a contradiction since f'"(x) is a non-zero constant when f(x) is a cubic polynomial. 15. With f(— 2) = 1 1 > and f(— 1) = — 1 < we conclude from the Intermediate Value Theorem that f(x) = x 4 + 3x + 1 has at least one zero between —2 and — 1. Then —2 < x < — 1 => — 8 < x 3 < — 1 => -32 < 4x 3 < -4 => -29 < 4x 3 + 3 < -1 => f'(x) < for -2 < x < -1 => f(x) is decreasing on [-2,-1] =>• f(x) = has exactly one solution in the interval (—2, —1). 16. f(x) = x 3 f'(x) = 3x 2 > on (—00, 0) =>• f(x) is increasing on (—00, 0). Also, f(x) < if x < — 2 and f(x) > if —2 < x < =>• f(x) has exactly one zero in (—00, 0). 17. g(t)= y/i+ 0+1-4 f® = * 2Vt+ j— > =>■ g(t) is increasing for t in (0, 00); g(3) = \/3 - 2 < and g(15) = v 15 > =>■ g(t) has exactly one zero in (0, 00). Cop # (c) 1 Pearson Ediratio^, k, publishing as Pearson Addison-Wesle 212 Chapter 4 Applications of Derivatives 18. g(t) = t^ + ^iTt - 3.1 => g'(t) = jjijj + ^= > => g(t) is increasing for t in (-1, 1); g(-0.99) = -2.5 and g(0.99) = 98.3 => g(t) has exactly one zero in (-1, 1). 19. r (0) = + sin 2 (§) - 8 =4> r'(0) = 1 + | sin (f) cos (f) = 1 + ± sin (f ) > on (-oo, oo) =4> r(0) is increasing on (— oo, oo); r(0) = —8 and r(8) = sin 2 (|) > => r(0) has exactly one zero in (— oo, oo). 20. r(0) = 20 - cos 2 + \[l =>• r'(0) = 2 + 2 sin cos = 2 + sin 20 > on (-oo, oo) =>■ r(0) is increasing on (-oo, oo); r(-27t) = -4?r - cos (-2tt) + \Jl = -4tt - 1 + \fl < and r(2?r) = 4tt-1 + \/2>0 => r(0) has exactly one zero in (— oo, oo). 21. r(0) = sec - i + 5 => r'(0) = (sec 0)(tan 0) + | > on (0, f ) => r(0) is increasing on (0, f ) ; r(O.l) ss -994 and r(1.57) « 1260.5 =4> r(0) has exactly one zero in (0, §) . 22. r(0) = tan - cot - =>• r'(0) = sec 2 + esc 2 - 1 = sec 2 + cot 2 > on (0, f ) on (0, § ) ; r (f ) = - \ < and r(1.57) « 1254.2 => r(0) has exactly one zero in (0, §) r(0) is increasing 23. By Corollary 1, f (x) = for all x =>■ f(x) = C, where C is a constant. Since f(— 1) = 3 we have C = 3 =4> f(x) = 3 for all x. 24. g(x) = 2x + 5 =>• g'(x) = 2 = f '(x) for all x. By Corollary 2, f(x) = g(x) + C for some constant C. Then f(0) = g(0) + C =>• 5 = 5 + C =» C = => f(x) = g(x) = 2x + 5 for all x. 25. g(x) = x 2 =>• g'(x) = 2x = f'(x) for all x. By Corollary 2, f(x) = g(x) + C. (a) f(0) = =5> = g(0) + C = + C => C = =>■ f(x) = x 2 =^ f(2) = 4 (b) f(l) = => = g(l) + C = 1+ C => C = -1 =>• f(x) = x 2 - 1 => f(2) = 3 (c) f(-2) = 3 => 3 = g(-2) + C => 3 = 4 + C =^> C = -1 =>■ f(x) = x 2 - 1 =>■ f(2) = 3 26. g(x) = mx => g'(x) = m, a constant. If f'(x) = m, then by Corollary 2, f(x) = g(x) + b = mx + b where b is a constant. Therefore all functions whose derivatives are constant can be graphed as straight lines y = mx + b. 27. (a) y = \ + C 28. (a) y = x 2 + C 29. (a) y' = -x~ 2 => y = i + C (b) y = f + C (b) y = x 2 - x + C (b) y = x+i+C (c) y = ^ + C (c) y = x 3 + x 2 - x + C (c) y = 5x-i+C 30. (a) y' = \ x- 1 / 2 => y = x 1 / 2 +C => y = v ^ + C (c) y = 2x 2 -2y / x + C 31. (a) y = - \ cos2t + C (c) y = - 1 cos 2t + 2 sin | + C (b) y = 2^+C (b) y = 2 sin \ + C 32. (a) y = tan + C (b) y' = 1 / 2 => y = § 3 / 2 + C 33. f(x) = x 2 - x + C; = f(0) = 2 - + C => C = => f(x) = x 2 - x (c) y = 2 6» 3 / 2 - tan + C Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 4.2 The Mean Value Theorem 213 34. g(x) = -i+x 2 + C;l=g(-l) 35. r(0) = 86* + cot 6 + C; = r (f ) : => r(8) = 8(9 + cot (9 - 2tt - 1 1 -^ + (-l) 2 + C =* C = -l => g(x) = -l+: (|) + cot (f ) + C =>• = 2tt + 1 + C =>• C = -2tt - 1 36. r(t) = sec t - t + C; = r(0) = sec (0) - + C =4> C = -1 =4> r(t) = sec t - t - 1 37. v 38. v 9.8t + 5 => s = 4.9t 2 + 5t + C; at s = 10 and t = we have C = 10 => s = 4.9t 2 + 5t + 10 32t - 2 =>• s = 16t 2 - 2t + C; at s = 4 and t = \ we have C = 1 => s = 6t 2 - 2t + 1 39. v = ^ = sin(7rt) => s = -icos(7rt) + C; at s = and t = we have C = \ =>• s 1 — COs(7Tt) 40. v = ^ = |cos(f ) =4> s = sin(|) + C; at s = 1 and t = 7r 2 we have C = 1 => s = sin(f ) + 1 41. a = 32 => v = 32t + d; at v = 20 and t = we have Ci = 20 => v = 32t + 20 =4> s = 16t 2 + 20t + C 2 ; at s = 5 and t = we have C 2 = 5 => s = 16t 2 + 20t + 5 42. a = 9.8 => v = 9.8t + Ci; at v = -3 and t = we have Ci = -3 => v = 9.8t - 3 => s = 4.9t 2 - 3t + C 2 ; at s = and t = we have C 2 = => s = 4.9t 2 - 3t 43. a = -4sin(2t) =4> v = 2cos(2t) + Ci; at v = 2 and t = we have Ci = => v = 2cos(2t) =>■ s = sin(2t) + C 2 ; at s = -3 and t = we have C 2 = —3 =^> s = sin(2t) — 3 44. a = 4cos(-) ^ v = -sin(-) + Ci; at v = and t = we have Ci = s = — 1 and t = we have C 2 = => s = — cosf — ) \ -sin( -) =^> s = -cos(-) + C 2 ; at -sinf-l 7T V 7T ^ 45. If T(t) is the temperature of the thermometer at time t, then T(0) = - 19° C and T(14) = 100° C. From the Mean Value Theorem there exists a < t < 14 such that T( ^lzl <0) = 8.5° C/sec = T'(t ), the rate at whic the temperature was changing at t = to as measured by the rising mercury on the thermometer. 46. Because the trucker's average speed was 79.5 mph, by the Mean Value Theorem, the trucker must have been going that speed at least once during the trip. 47. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that speed at least once during the trip. 48. The runner's average speed for the marathon was approximately 1 1.909 mph. Therefore, by the Mean Value Theorem, the runner must have been going that speed at least once during the marathon. Since the initial speed and final speed are both mph and the runner's speed is continuous, by the Intermediate Value Theorem, the runner's speed must have been 1 1 mph at least twice. 49. Let d(t) represent the distance the automobile traveled in time t. The average speed over < t < 2 is d(2 ^~p (0) . The Mean Value Theorem says that for some < t < 2, d'(t ) the speed of the automobile at time t (which is read on the speedometer) d( ^:Q (0) . The Mean Value Theorem says that for some < t < 2, d'(t ) = ^I^ - The value d'(t ) is 50. a(t) = v'(t) = 1.6 => v(t) = 1.6t + C; at (0, 0) we have C = =>■ v(t) = 1.6t. When t = 30, then v(30) = 48 m/sec. Cop # (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 214 Chapter 4 Applications of Derivatives 5 1 . The conclusion of the Mean Value Theorem yields b-a 3 => ^ 2 (^)=a-b =* c ab. 52. The conclusion of the Mean Value Theorem yields \_\ = 2c => c — '' 2 ' 53. f'(x) = [cos x sin(x + 2) + sin x cos(x + 2)] — 2 sin(x + 1) cos(x + 1) = sin(x + x + 2) — sin 2(x + 1) = sin (2x + 2) — sin (2x + 2) = 0. Therefore, the function has the constant value f(0) = —sin 2 1 w —0.7081 which explains why the graph is a horizontal line. 54. (a) f(x) = (x + 2)(x + l)x(x - l)(x - 2) = x 5 - 5x 3 + 4x is one possibility. (b) Graphing f(x) = x 5 - 5x 3 + 4x and f'(x) = 5x 4 - 15x 2 + 4 on [-3, 3] by [-7, 7] we see that each x-intercept of f'(x) lies between a pair of x-intercepts of f(x), as expected by Rolle's Theorem, y y = f'(x) y = f(x) (c) Yes, since sin is continuous and differentiable on ( — oo, oo). 55. f(x) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f(x) is zero twice between a and b. Then by the Mean Value Theorem, f'(x) would have to be zero at least once between the two zeros of f(x), but this can't be true since we are given that f'(x) ^ on this interval. Therefore, f(x) is zero once and only once between a and b. 56. Consider the function k(x) = f(x) — g(x). k(x) is continuous and differentiable on [a, b], and since k(a) = f(a) — g(a) and k(b) = f(b) — g(b), by the Mean Value Theorem, there must be a point c in (a, b) where k'(c) = 0. But since k'(c) = f'(c) — g'(c), this means that f'(c) = g'(c), and c is a point where the graphs of f and g have tangent lines with the same slope, so these lines are either parallel or are the same line. 57. Yes. By Corollary 2 we have f(x) = g(x) + c since f'(x) = g'(x). If the graphs start at the same point x = a, then f(a) = g(a) =4> c = =>- f(x) = g(x). 58. Let f(x) = sin x for a < x < b. From the Mean Value Theorem there exists a c between a and b such that sin b — sin a b-a cose =>■ -1< sin £- sina < 1 => — b — a — I si "b-a" a | ^ l =*" |sinb-sina| < |b-a|. 59. By the Mean Value Theorem we have ( g _ a = f'(c) for some point c between a and b. Since b — a > and f(b) < f(a), we have f(b) - f(a) < => f'(c) < 0. 60. The condition is that f should be continuous over [a, b]. The Mean Value Theorem then guarantees the existence of a point c in (a, b) such that ^ ~ a = f '(c). If f ' is cont maximum value on [a, b], and min f < f'(c) < max f ', as required. existence of a point c in (a, b) such that ^ _ a = f '(c). If f ' is continuous, then it has a minimum and Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 4.3 Monotonic Functions and the First Derivative Test 215 61. f'(x) = (1 + x 4 cos x) * => f"(x) = - (1 + x 4 cos x) 2 (4x 3 cos x - x 4 sin x) -x 3 (1 + x 4 cos x) (4 cos x — x sin x) < for < x < 0.1 => f'(x) is decreasing when < x < 0.1 Mi o.i min f « 0.9999 and max f = 1. Now we have 0.9999 < f(0 -' ) ~ 1 < 1 => 0.09999 < f(0.1) - 1 < 0.1 => 1.09999 < f(0.1) < 1.1. 62. f (x) = (1 - x 4 ) -1 => f"(x) = - (1 - x 4 )~ 2 (-4x 3 ) = -^-3 > for < x < 0.1 => f'(x) is increasing when (1 X ) f(0.1)-2 0.1 < f(0.1) - 2 < 0.10001 =*> 2.1 < f(0.1) < 2.10001. < x < 0.1 => min f = 1 and max f = 1.0001. Now we have 1 < ' ^ < 1.0001 63. (a) Suppose x < 1, then by the Mean Value Theorem f(x ^I^ (1) < => f(x) > f(l). Suppose x > 1, then by the Mean Value Theorem fJ ^zj 11 > =>■ f(x) > f(l). Therefore f(x) > 1 for all x since f(l) = 1. (b) Yes. From part (a), lim f(x) ^ (1) < and lim fiMnIil > q. Since f'(l) exists, these two one-sided X — > 1~ X ' X — > 1+ x limits are equal and have the value f'(l) =4> f'(l) < and f'(l) > ^ f'(l) = 0. 64. From the Mean Value Theorem we have ( ^ ~ a (a) = f'(c) where c is between a and b. But f'(c) = 2pc + q = has only one solution c = — £- . (Note: p ^ since f is a quadratic function.) 4.3 MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST 1. (a) f'(x) = x(x — 1) =>- critical points at and 1 (b) f = +++ | | +++ => increasing on (— oo, 0) and (1, oo), decreasing on (0, 1) 1 (c) Local maximum at x = and a local minimum at x = 1 2. (a) f'(x) = (x - l)(x + 2) => critical points at -2 and 1 (b) f = +++ | | +++ => increasing on (— oo, —2) and (1, oo), decreasing on (—2, 1) -2 1 (c) Local maximum at x = —2 and a local minimum at x = 1 3. (a) f'(x) = (x - l) 2 (x + 2) =>- critical points at -2 and 1 (b) f' = | +++ | +++ => increasing on (—2. 1) and (1, oo), decreasing on (—oo, —2) -2 1 (c) No local maximum and a local minimum at x = —2 4. (a) f'(x) = (x - l) 2 (x + 2) 2 =^> critical points at -2 and 1 (b) f = +++ | +++ | +++ => increasing on (— oo, —2) U (—2, 1) U (1, oo), never decreasing -2 1 (c) No local extrema 5. (a) f'(x) = (x - l)(x + 2)(x - 3) => critical points at -2, 1 and 3 (b) f = | +++ | | +++ => increasing on (—2, 1) and (3, oo), decreasing on (— oo, —2) and (1,3) -2 13 (c) Local maximum at x = 1, local minima at x = —2 and x = 3 6. (a) f'(x) = (x - 7)(x + l)(x + 5) =^ critical points at -5, -1 and 7 (b) f = | +++ | I +++ => increasing on (—5, —1) and (7, oo), decreasing on (—oo, —5) and (—1,7) -5 -17" (c) Local maximum at x = — 1, local minima at x = —5 and x = 7 Cop # (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 216 Chapter 4 Applications of Derivatives 7. (a) f'(x) = x~ 1/3 (x + 2) =4> critical points at -2 and (b) f = +++ | )( +++ =>• increasing on (— oo, —2) and (0, oo), decreasing on (—2, 0) -2 (c) Local maximum at x = —2, local minimum at x = 8. (a) f (x) = x~ 1/2 (x - 3) =>■ critical points at and 3 (b) f = ( | +++ =>■ increasing on (3, oo), decreasing on (0, 3) 3 (c) No local maximum and a local minimum at x = 3 9. (a) g(t) = -t 2 - 3t + 3 => g'(t) = -2t - 3 => a critical point at t 3 . _/ (— oo, — |) , decreasing on (— |, oo) (b) local maximum value of g (— |) - (c) absolute maximum is =j- at t = — | (d) g(t) g (t) = -t 2 -3t + 3 6- f att=-| -+ | , increasing on -3/2 10. (a) g(t) = -3t 2 + 9t + 5 =>• g'(t) = -6t + 9 => a critical point at t = | ; g' = ++- (— oo, |) , decreasing on (|, oo) (b) local maximum value ofg(|) = ^ att (c) absolute maximum is ^ at t = | (d) g(t) = -3t 2 +9t + 5 3/2 increasing on 11. (a) h(x)= -x 3 + 2x 2 => h' = 4/3 (b) local maximum value of h ( j) (c) no absolute extrema h'(x) = -3x 2 + 4x = x(4 - 3x) =^ critical points at x = 0, \ +++ | , increasing on (0, |) , decreasing on (— oo,0) and (|, Yj at x = | ; local minimum value of h(0) = at x = Copffigl (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle Section 4.3 Monotonic Functions and the First Derivative Test 217 (d) x 3 +2x 2 12. (a) h(x) = 2x 3 - 18x =>■ h'(x) = 6x 2 - 18 = 6 be + y/3) (x - y/3) => critical points at x = ± \/3 =>• h' = +++ I I +++, increasing on ( —00, — y 3 ) and ( y3, 00 ) , decreasing on ( — y3, y 3 -n/3 y^ V y V ' V (b) a local maximum is h ( — y 3 I = 12 y 3 at x = — y 3; local minimum is h ( y 3 J = — 12y 3 at x = y 3 (c) no absolute extrema (d) 2 3 h(x) = 2x 3 -18x 13. (a) f(0) = 36 2 - 46 3 => f (0) = 66- 126 2 = 60(1 - 20) => critical points at 6 = 0, \ => f increasing on (0, |) , decreasing on (—00, 0) and (|, 00) (b) a local maximum is f (|) = | at = |, a local minimum is f(0) = at 6* = (c) no absolute extrema (d) f(0) = 30 2 -40 3 I+++I - 1/2 14. (a) f(0) = 60 - 3 => f'(0) = 6 — 36> 2 = 3 (y^ - 0) (y 7 ^ + 0) =>■ critical points at = ± y^ => f ' = I +++ I , increasing on ( — y 2, y 2), decreasing on ( — 00, — y 2) and ( y 2, 00) -V2 V2 (b) a local maximum is f f y 2 J = 4y 2 at 6 = y 2, a local minimum is f f — y 2 j = — 4y 2 at 6 = — y 2 (c) no absolute extrema Cop # (c) 1 Pearson Education, It, publishing as Pearson Addison-Wesle 218 Chapter 4 Applications of Derivatives (d) f(e) f(e) = 6e-e 3 ♦*- e 15. (a) f(r) = 3r 3 + 16r =>• f'(r) = 9r 2 + 16 =4> no critical points decreasing (b) no local extrema (c) no absolute extrema (d) f(r) 200 ; / f(r) = 3r ! + 16r f = +++++, increasing on (— oo, oo), never 16. (a) h(r) = (r + if => h'(r) = 3(r + 7) 2 => a critical point at r = -7 =$■ h' (— oo, —7) U (—7, oo), never decreasing (b) no local extrema (c) no absolute extrema (d) h(r) 10 4 h(r) = (r + 7) 3 -++ | +++, increasing on -7 -5 -10 17. (a) f(x) = x 4 - 8x 2 + 16 => f (x) = 4x 3 - 16x = 4x(x + 2)(x - 2) =*• critical points at x = and x = ±2 =^ f ' = | +++ | | +++, increasing on (—2, 0) and (2, oo), decreasing on (— oo, —2) and (0, 2) -2 2 (b) a local maximum is f(0) = 16 at x = 0, local minima are f(±2) = 0atx= ±2 (c) no absolute maximum; absolute minimum is at x = ±2 (d) f(x) f (x) = x 4 -8x 2 + 16 Cop # (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Section 4.3 Monotonic Functions and the First Derivative Test 219 18. (a) g(x) = x 4 - 4x 3 + 4x 2 => g'(x) = 4x 3 - 12x 2 + 8x = 4x(x - 2)(x - 1) =>■ critical points at x = 0, 1, 2 =>• g' = | +++ | | +++, increasing on (0, 1) and (2. oo), decreasing on (— oo,0) and (1, 2) 12 (b) a local maximum is g(l) = 1 at x = 1, local minima are g(0) = at x = and g(2) = at x = 2 (c) no absolute maximum; absolute minimum is at x = 0, 2 (d) g(x) = x 4 -4x J +4x 2 19. (a) H(t) = 1 1 4 — t G => H'(t) = 6t 3 - 6t 5 = 6t 3 (l + t)(l - t) =>■ critical points at t = 0, ±1 => H' = +++ | | +++ | , increasing on (— oo, —1) and (0, 1), decreasing on (— 1, 0) and (1, oo) -10 1 (b) the local maxima are H(— 1) = \ at t = — 1 and H(l) = \ at t = 1, the local minimum is H(0) = at t = (c) absolute maximum is | at t = ± 1 ; no absolute minimum (d) H(t) 1 1 - H(t) = ft 4 -t 6 20. (a) K(t) = 15t 3 - t 5 =4> K'(t) = 45t 2 - 5t 4 = 5t 2 (3 + t)(3 - t) => critical points at t = 0, ±3 => K' = | +++ | +++ | , increasing on (—3,0) U (0, 3), decreasing on (— oo, —3) and (3, oo) -3 3 (b) a local maximum is K(3) = 162 at t = 3, a local minimum is K(— 3) = —162 at t = — 3 (c) no absolute extrema (d) K < t > K(t) = 15t'-t 5 150 ■ 100 21. (a) g(x) = xy/i x 2 ) 1/2 => g'(x) = (8-x 2 ' 1/2 +*g; x 2 )" 1/2 (-2x) -2V2 , -2] and ( 2,2V2 2(2 - x)(2 + x) 'isjl -x) (2^/2 + > => critical points at x = ±2, ± l^fl => g' = ( | +++ | -2^ - 2 2 ) , increasing on (—2, 2), decreasing on 2a/2 Cop # (c) 1 Pearson Ediratio^, k, publishing as Pearson Addison-Wesle 220 Chapter 4 Applications of Derivatives (b) local maxima are g(2) = 4 at x = 2 and g ( — 2 \J 2 ) = at x = — 2 y 2, local minima are g(— 2) = —4 at x = -2 and g I 2\/2 J = at x = 2V2 (c) absolute maximum is 4 at x = 2; absolute minimum is —4 at x = —2 (d) g(x) (x) = xV8-x 2 2V5-X , increasing on (0, 4), decreasing on (— oo, 0) 22. (a) g(x) = xV5-x = x 2 (5 - x) 1 / 2 => g'(x) = 2x(5 - x) 1 / 2 + x 2 (±) (5 - x)- 1/2 (-l) - 5xl "' ' ! => critical points at x = 0, 4 and 5 =>• g' = | + 4 5 and (4, 5) (b) a local maximum is g(4) = 1 6 at x = 4, a local minimum is at x = and x = 5 (c) no absolute maximum; absolute minimum is at x = 0, 5 (d) ;(x) = x 2 V5-x 23. (a) f(x) x 2 -3 x-2 f'(x) 2x(x-2)-(x 2 -3)(l) _ (x-3)(x-l) (x - 2) 2 (x - 2) 2 f -+- 2 +- critical points at x = 1,3 increasing on (— oo, 1) and (3, oo), decreasing on (1, 2) and (2, 3), 12 3 discontinuous at x = 2 (b) a local maximum is f(l) = 2 at x = 1, a local minimum is f(3) = 6 at x = 3 (c) no absolute extrema (d) f(x) B . 6 4 2 f (x) = v ; x-2 /<?2 \2 4 6 24. (a) f(x) = 3^ => f = ++- el, x 3x 2 (3x 2 + l)-x 3 (6x) 3x 2 (x 2 + l) ... , ... A f (x) = (3x 2 + i) 2 = (3x 2 + i) 2 => a cntlcal P° lnt at x = -+, increasing on (— oo, 0) U (0, oo), and never decreasing (b) no local extrema (c) no absolute extrema Cop # (c) 1 Pearson Ediratio^, k, publishing as Pearson Addison-Wesle Section 4.3 Monotonic Functions and the First Derivative Test 221 (d) -0.5 25. (a) f(x) = x^ 3 (x + 8) = x 4 / 3 + 8X 1 / 3 => f ( x ) = f x 1 / 3 + f x~ 2 / 3 = ^±^ =>■ critical points at x = 0, -2 =^ f = | +++)(+++, increasing on (—2, 0) U (0, oo), decreasing on (— oo, —2) -2 (b) no local maximum, a local minimum is f(— 2) = —6 V 2 w —7.56 at x = —2 (c) no absolute maximum; absolute minimum is —6 y 2 at x = —2 (d) f(x) 10 + -10 f(x) = x" 3 (x + 8) 26. (a) g(x) = x 2 / 3 (x + 5) = x 5 / 3 + 5x 2 / 3 => g'(x) = f x 2 / 3 + ^ x" 1 / 3 = 5^ + 2) ^> critical points at x = -2 and x = =^ g' = +++ | )(+++, increasing on (— oo, —2) and (0, oo), decreasing on (—2,0) -2 (b) local maximum is g(— 2) = 3 y 4 w 4.762 at x = —2, a local minimum is g(0) = at x = (c) no absolute extrema (d) 9(1) x + 5 12 3 27. (a) h(x) = x 1 / 3 (x 2 - 4) = x 7 / 3 - 4X 1 / 3 => h'(x) = \ x 4 / 3 - f x~ 2 / 3 /7x + 2 K/7x-2 3^ critical points at x -0 i 2 - h' - + - )( -+ -2/\/7 2/^7 increasing on f — oo, —h ) and ( -?-, oo J , decreasing on (^,0)and(0,^) / —9 \ 24 \/2 _9 / 9 \ 24 \/2 (b) local maximum is h I — f- I = T ' 6 « 3.12 at x = —£, the local minimum is h I -4- 1 = (c) no absolute extrema 7V6 ^3.12 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 222 Chapter 4 Applications of Derivatives (d) h(x) = x"' 3 (x 2 -4) 28. (a) k(x) = x 2 / 3 (x 2 - 4) = x 8 / 3 - 4x 2 / 3 => k'(x) = § x 5 / 3 - | x" 1 / 3 = 8(x+ 3 1 3 ^" 1) =^> critical points at x = 0, ± 1 =>- k' = | +++)( | +++, increasing on (— 1, 0) and (1, oo), decreasing on (— oo. — 1) -10 1 and (0,1) (b) local maximum is k(0) = at x = 0, local minima are k ( ± 1) = — 3 at x = ±1 (c) no absolute maximum; absolute minimum is — 3 at x = ±1 (d) k(x) = x 2 ' 3 (x 2 -4) 29. (a) f(x) = 2x - x 2 =>• f (x) = 2 - 2x = 2(1 - x) =>• a critical point at x = 1 => f = +- f(2) = => a local maximum is 1 at x = 1, a local minimum is at x = 2 (b) absolute maximum is 1 at x = 1 ; no absolute minimum (c) f (x) = 2x-x 2 -] andf(l)= 1, 2 30. (a) f(x) = (x + l) 2 => f (x) = 2(x + 1) => a critical point at x = -1 =>• f = | +++ ] and f(-l) = 0, f(0) = 1 -1 =>• a local maximum is 1 at x = 0, a local minimum is at x = — 1 (b) no absolute maximum; absolute minimum is at x = — 1 (c) f(x) \ 5- \f(x) = (x + l) 2 4 - \ 3 - \ 2 - \^ 1, -4 -3 -2 -1 -1 1 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 4.3 Monotonic Functions and the First Derivative Test 223 31. (a) g(x) = x 2 - 4x + 4 =>• g'(x) = 2x - 4 = 2(x - 2) => a critical point at x = 2 =>• g' = [ | +++ and 12 g(l) = 1, g(2) = =^ a local maximum is 1 at x = 1, a local minimum is g(2) = at x = 2 (b) no absolute maximum; absolute minimum is at x = 2 (c) g(x) S' 4 ■ 3 - g(x) = x : -4x + 4 2 - 1 ■ 1 1 2 3 4 32. (a) g(x) = -x 2 - 6x - 9 =^ g'(x) = -2x - 6 = -2(x + 3) =4> a critical point at x = -3 = [ +- -4 g(— 4) = — 1, g(— 3) = =>- a local maximum is at x = —3, a local minimum is — 1 at x = —4 (b) absolute maximum is at x = —3; no absolute minimum (c) g(x) -3 and -10 33. (a) f(t) = 12t - t 3 => f'(t) = 12 - 3t 2 = 3(2 + t)(2 - t) => critical points at t = ±2 =>• f' = [ | - -3 -2 and f(-3) = -9, f(-2) = - 16, f(2) = 16 => local maxima are -9 at t = -3 and 16 at t = -2, a local minimum is — 16 at t = —2 (b) absolute maximum is 16 at t = 2; no absolute minimum (c) f(t) f(t) = 12t-t 3 - + 34. (a) f(t) = t 3 - 3t 2 => f '(t) = 3t 2 - 6t = 3t(t - 2) =>• critical points at t = and t = 2 => f = +++ | | +++ ] and f(0) = 0, f(2) = -4, f(3) = =^ a local maximum is at t = and t = 3, a 2 3 local minimum is —4 at t = 2 (b) absolute maximum is at t = 0, 3; no absolute minimum Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 224 Chapter 4 Applications of Derivatives (c) f(t) ♦- 1 35. (a) h(x) 2x 2 + 4x =>• h'(x) = x 2 - 4x + 4 = (x - 2) 2 =4> a critical point at x = 2 h' = [ +- -++ and h(0) = => no local maximum, a local minimum is at x = (b) no absolute maximum; absolute minimum is at x = (c) h(x) 6 5 4 3 2 1 h(x) = 2x 2 +4x v ' 3 36. (a) k(x) = x 3 + 3x 2 + 3x + 1 =>■ k'(x) = 3x 2 + 6x + 3 = 3(x + l) 2 => a critical point at x = -1 =>■ k' = +++ | +++ ] and k(— 1) = 0, k(0) = 1 =$■ a local maximum is 1 at x = 0, no local minimum -1 (b) absolute maximum is 1 at x = 0; no absolute minimum (c) k(x) 2 k(x) = x 3 +3x 2 +3x+l 1 37. (a) f(x)= |-2sin(|; f 2tt/3 ^ f (x) = i - cos (I) , f (x) = => cos (I) = \ =4> a critical point at x = y -++ ] and f(0) = 0, f (?f ) = f - v/3, f(27r) = ?r ^> local maxima are at x = and tt 2tt at x = 27r, a local minimum is | — y 3 at x 2^ (b) The graph of f rises when f > 0, falls when f ' < 0, and has a local minimum value at the point where f ' changes from negative to positive. Cop # (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Section 4.3 Monotonic Functions and the First Derivative Test 225 „2 . 38. (a) f(x) = —2 cos x — cos^ x X = -7T, 0, 7T => f =>■ f (x) = 2 sin x + 2 cos x sin x = 2(sin x)(l + cos x) =>■ critical points at | +++ ] and f(— 7r) = 1, f(0) = —3, f(7r) = 1 => a local maximum is 1 at — 7T Q 7T x = ± 7T, a local minimum is —3 at x = (b) The graph of f rises when f ' > 0, falls when f ' < 0, ¥ and has local extreme values where f ' = 0. The function f has a local minimum value at x = 0, where the values of f ' change from negative to positive. /(x) = -2cos;r -cos 2 *, -it < x < ir 39. (a) f(x) = esc 2 x - 2 cot x => f '(x) = 2(csc x)(-csc x)(cot x) - 2 (-esc 2 x) = -2 (esc 2 x) (cot x - 1) a critical point at x => f = (— I + tt/4 n (b) The graph of f rises when f ' > 0, falls when f ' < 0. and has a local minimum value at the point where f ' = and the values of f ' change from negative to positive. The graph of f steepens as f'(x) — > ± oo ) and f (f ) =0 =>• no local maximum, a local minimum is at x fix) = csc ! jr -2cotx, < x < n 40. (a) f(x) = sec 2 x — 2 tan x => f'(x) = 2(sec x)(sec x)(tan x) — 2 sec 2 x = (2 sec 2 x) (tan x — 1) =4> a critical point atx=|^f' = ( | +++) andf(f) = -7t/2 tt/4 tt/2 (b) The graph of f rises when f ' > 0, falls when f ' < 0, and has a local minimum value where f ' = and the values of f ' change from negative to positive. no local maximum, a local minimum is at x f(x) = sec 2 jr -2 tan* 41. h(0) = 3 cos (I) => h'(0) = - | sin (§) =^ h' = [ ] , (0,3) and (2tt, -3) =^ a local maximum is 3 at 9 = 0, 2tt a local minimum is — 3 at 6 = 2ir 42. h(0) = 5 sin (f ) => h'(0) = | cos (f ) =>• h' = [ +++ ] , (0, 0) and (tt, 5) => a local maximum is 5 at 8 = n, a local minimum is at 9 = Copffigl (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 226 Chapter 4 Applications of Derivatives 43. (a) y =/M r\ lal (b) y =m 1 (c) (c) (d) i (d) 44. (a) y - f(x) (b) (c) (d) fix! 45. (a) y 1 . s / 2 ****/ N**-*- y = g(x) 2 (b) 46. (a) , I 1- f y = h(x) -t -J o 2 4 47. f(x) = x 3 - 3x + 2 => f (x) = 3x 2 - 3 = 3(x - l)(x + 1) => f = +++ f (x) > for x = c = 2. +++ =4> rising for x = c = 2 since f(x) = ax 2 + bx + c = a (x ; (x 2 + ^) + c = a(x 2 + fex+^)-£ + c = a(x+^) b \ 2 b 2 - 4ac a parabola whose 4a 2 J 4a ' ""T ' 2ai 4a vertex is at x = — ^ . Thus when a > 0, f is increasing on (^, oo) and decreasing on (—00, ^) ; when a < 0, f is increasing on (—00. =^) and decreasing on (=£, 00) . Also note that f'(x) = 2ax + b = 2a (x + |-) => for a>0, f' = I +++ ; for a < 0, f = +++ | . — b/2a -b/2a 4.4 CONCAVITY AND CURVE SKETCHING 1. y=^-^-2x+i^>y' = x 2 -x-2 = (x- 2)(x +1) => y" = 2x - 1 = 2 (x - |) . The graph is rising on (—00, —1) and (2, 00), falling on (— 1, 2), concave up on (|, 00) and concave down on (—00, |) . Consequently, a local maximum is | at x = — 1 , a local minimum is — 3 at x = 2, and ( h , — |) is a point of inflection. Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 4.4 Concavity and Curve Sketching 227 2. y = si - 2x 2 + 4 => y' = x 3 - 4x = x (x 2 - 4) = x(x + 2)(x - 2) => y" = 3x 2 - 4 = f y^x + 2^ (y^x - 2) . The graph is rising on (—2, 0) and (2, oo), falling on (— oo, —2) and (0, 2), concave up on ( — oo, -y- j and I -4= , oo 1 and concave down on ( — 7" ) ~7~ ) • Consequently, a local maxirr x = ± 2, and ( — j- , ^ j and ( -4= , y J are points of inflection. -j- j -7- ) • Consequently, a local maximum is 4 at x = 0, local minima are at 3. y = ! (x 2 - 1) 2/3 => y'=(|)(|)(x 2 -l)- 1/3 (2x) = x(x 2 -l)- 1/3 ,y' = -— ) (+++ | —)(+++ -10 1 =>■ the graph is rising on (—1, 0) and (1, oo), falling on (— oo, —1) and (0, 1) =^ a local maximum is | at x = 0, local minima are at x = ± 1; y" = (x 2 - 1) ' + (x) (- i) (x 2 - 1) ' (2x) \-^f x 2 -3 3 V(" 2 -!) 4 + + + -) ( -1 )(• 1 - | +++ =4> the graph is concave up on ( — oo, — y 3 J and ( y 3, oo j , concave down on I — y 3, y 3 I => points of inflection at ( ± y 3, -j- j 4. y = 9. X V3 (x 2 _ 7) ^ y'= t L x - 2 / 3 (x 2 -7) + ^x 1 /3(2x)=|x- 2 / 3 (x 2 -l),y' = +++| )( |+++ =>■ the graph is rising on (— oo, —1) and (1, oo), falling on (—1, 1) => a local maximum is y at x = — 1, a local minimum is - y atx = 1; y" = -x~ 5 / 3 (x 2 - 1) + 3X 1 / 3 = 2x : / 3 + x~ 5 / 3 = x~ 5 / 3 (2x 2 + 1) , y" = )( +++ =4> the graph is concave up on (0, oo), concave down on (— oo, 0) => a point of inflection at (0,0) 5. y = x + sin 2x => y' = 1 + 2 cos 2x, y' = [ | +++ | ] => the graph is rising on (— |, |) , falling -2tt/3 -tt/3 tt/3 2tt/3 ^ and z _l YJ! f,-f)and(f,?f local maxima are — y + -y- at x at x = | , local minima are 3 2 at X — 3 """ 3 2 5 and ^ - ^ at x = y" ; y" = -4 sin 2x, y" = [ 2tt/3 -tt/2 - | +++ ] =>■ the tt/2 2tt/3 graph is concave up on (— |, 0) and (|, y) , concave down on (— y, — |) and (0, |) => points of inflection at (-f,-f), (0,0), and (|, |) 6. y = tan x — 4x =S> y' = sec 2 x — 4, y' = ( +++ | | +++ ) => the graph is rising on (— |, — |) and -tt/2 -tt/3 tt/3 tt/2 (|, |) , falling on (— f , f ) => a local maximum is — v 3 + y atx = — | , a local minimum is y 3 - j at x = |; y" = 2(sec x)(sec x)(tan x) = 2 (sec 2 x) (tan x), y" = ( | +++ ) =^ the graph is concave up on (0, |) , -tt/2 tt/2 concave down on (— | , 0) => a point of inflection at (0, 0) 7. If x > 0, sin |x| = sin x and if x < 0, sin |x| = sin(— x) = —sin x. From the sketch the graph is rising on (- f , - § ) , (0, f ) and (f , 2tt) , falling on (-2tt, - 3 f ) , (— |,0) and (|, y) ; local minima are —1 at x = ± y and at x = 0; local maxima are 1 at x = ± | and at x = ± 2tt; concave up on (— 2ir, — n) and (ir, 2ir), and concavedown on {—it, 0) and (0, ir) =$■ points of inflection are (— n, 0) and (w, 0) (-2R.0) (-n/2, 1) (-37C/2.-1) y = sin|jr|, -2ir sjs2t (*/2.D (27t,0) (3jc/2,-1) Cfiojt (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 228 Chapter 4 Applications of Derivatives y = 2 cos x — v2) -2 sin x — v 2, y' = [ — 7T 3n; 4 ' atx y" = +++ | I +++] =>■ rising on ■3tt/4 -tt/4 5tt/4 3tt/2 |) and (^, ^) , falling on (-tt, - 2s) and (- f , ^ ) => local maxima are -2 + tta/2 at x = -tt, y^ + ^ 4 """ 2 2cosx, y" = [ +++ I I +++] -7r -tt/2 tt/2 3tt/2 and — :-^ al x 4?, and local minima are - \[7. + ^ at x = _ 3| and _ ^ _ 5^ at x = ^ . 7T 7T ' 2' 2; concave up on (— it, — |) and (|, y) , concave down on points of inflection at('-f,^)and('f,-^f E ) 9. When y = x 2 - 4x + 3, then y' = 2x - 4 = 2(x - 2) and y" = 2. The curve rises on (2, oo) and falls on (— oo, 2). At x = 2 there is a minimum. Since y' > 0, the curve is concave up for all x. >' V' \ J _ 1 2 1 -4 -3 -2 -1 l\ 2 /3 4 -1 (2,-1) -2 Locmin 10. When y = 6 - 2x - x 2 , then y' = -2 - 2x = -2(1 + x) and y" = —2. The curve rises on (-co, —1) and falls on (— 1 , oo). At x = — 1 there is a maximum. Since y' < 0, the curve is concave down for all x. 11. When y = x 3 - 3x + 3, then y' = 3x 2 - 3 = 3(x - l)(x + 1) and y" = 6x. The curve rises on (— oo, — 1) U (1, oo) and falls on (—1, 1). At x = — 1 there is a local maximum and at x = 1 a local minimum. The curve is concave down on (— oo, 0) and concave up on (0, oo). There is a point of inflection at x = 0. Loc y = x i --ix + i max ( - l - S W 5 \ / V I / Inf 1 1 / 2 / 1 ' V d.l) / Loc min 1 1 1 12. When y = x(6 - 2x) 2 , then y' = -4x(6 - 2x) + (6 - 2x) 2 = 12(3 - x)(l - x) and y" = -12(3 - x) - 12(1 - x) = 24(x — 2). The curve rises on (— oo, 1) U (3, oo) and falls on (1, 3). The curve is concave down on (— oo, 2) and concave up on (2, oo). At x = 2 there is a point of inflection. Copyright (c) 2006 Pearson Education Section 4.4 Concavity and Curve Sketching 229 13. When y = -2x 3 + 6x 2 - 3, then y' = -6x 2 + 12x = -6x(x - 2) and y" = -12x + 12 = -12(x - 1). The curve rises on (0, 2) and falls on (— oo, 0) and (2, oo). At x = there is a local minimum and at x = 2 a local maximum. The curve is concave up on (— oo, 1) and concave down on (1, oo). At x = 1 there is a point of inflection. (2, 5) Loc max y = -2^ + &!''- 3 14. When y = 1 - 9x - 6x 2 - x 3 , then y' = -9 - 12x - 3x 2 = -3(x + 3)(x + 1) and y" = -12 - 6x = -6(x + 2). The curve rises on (—3, —1) and falls on (— oo, —3) and (— 1, oo). At x = — 1 there is a local maximum and at x = — 3 a local minimum. The curve is concave up on (— oo, —2) and concave down on (—2, oo). At x — —2 there is a point of inflection. y = 1 - 9x - 6jc 2 - x 15. When y = (x - 2) 3 + 1, then y' = 3(x - 2) 2 and y" = 6(x — 2). The curve never falls and there are no local extrema. The curve is concave down on (— oo, 2) and concave up on (2, oo). At x = 2 there is a point of inflection. y / i 3 / 2 1 Infl / (2,1)/ -2 - 1 /l 2 3 4 -1 " / -2 / ). = ( J :-2) 3 +l 16. When y = 1 - (x + l) 3 , then y' = -3(x + l) 2 and y" = — 6(x + 1). The curve never rises and there are no local extrema. The curve is concave up on (— oo, —1) and concave down on (—1, oo). At x = 1 there is a point of inflection. Copyright (c) 2006 Pearson Education 230 Chapter 4 Applications of Derivatives 17. When y = x 4 - 2x 2 , then y' = 4x 3 - 4x = 4x(x + l)(x - 1) and y" = 12x 2 - 4 = 12 (x + 4?) [x - 4- J . The curve rises on (—1,0) and (1, oo) and falls on (— oo, —1) and (0, 1). At x = ±1 there are local minima and at x = a local maximum. The curve is concave up on ( — oo, \-\ and I -T- , oo ) and concave down on there are points of inflection. v^'vO Atx =7I t y = x 4 -2x 2 \ - 1 / i Loc max f \ (0, 0) v / -2 \-l J V 1 / 2 Loc min \ /> (-1.-D ^7 (-1/^,-5/9) Infl ^^ yLoc min - I^O.-I) fl/V3,-5/9) Infl 18. When y = -x 4 + 6x 2 - 4, then y' = -4x 3 + 12x = -4x (x + v 7 ^) U - V^) and y" = - 12x2 + 12 = — 12(x + l)(x — 1). The curve rises on ( —00, — y 3] and (0, \/3) , and falls on (-\/3, 0] and (\/3, 00] . At x = ± v 3there are local maxima and at x = a local minimum. The curve is concave up on (—1, 1) and concave down on (—00, — l)and (1, 00). At x = ±1 there are points of inflection. (- ^".5) (/5,5) y = -x' + dx 1 - 4 (0.-4) 19. When y = 4x 3 - x 4 , then y' = 12x 2 - 4x 3 = 4x 2 (3 - x) and y" = 24x — 12x 2 = 12x(2 — x). The curve rises on (—00, 3) and falls on (3, 00). At x = 3 there is a local maximum, but there is no local minimum. The graph is concave up on (0, 2) and concave down on (—00, 0) and (2, 00). There are inflection points at x = and x = 2. > k (3, 27) 11 Loc max 21 1(2, 16)/ I y = 4x 3 -x 4 15 ~_ flnfl 9 Infl (0,0) 3_ (\ .12 3 I 4 20. When y = x 4 + 2x 3 , then y' = 4x 3 + 6x 2 = 2x 2 (2x + 3) and y" = 12x 2 + 12x = 12x(x +1). The curve rises on (— |, 00) and falls on (—00, — |) . There is a local minimum at x = — |, but no local maximum. The curve is concave up on (—00, —1) and (0, 00), and concave down on (—1,0). At x = — 1 and x = there are points of inflection. Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle Section 4.4 Concavity and Curve Sketching 231 21. When y = x 5 - 5x 4 , then y' = 5x 4 - 20x 3 = 5x 3 (x - 4) and y" = 20x 3 — 60x 2 = 20x 2 (x — 3). The curve rises on (— oo, 0) and (4, oo), and falls on (0, 4). There is a local maximum at x = 0, and a local minimum at x = 4. The curve is concave down on (— oo, 3) and concave up on (3, oo). At x = 3 there is a point of inflection. 22. When y = x = (|-5 •5' 3 (f 3 (1) then y' , and y" -5 x(4)(|-5) 3 (i '5x v 2 5) 2 (x \2 ") \2> ,■ , y ,. _, ,. M ., ... -J ,.* 4). The curve is rising on (— oo, 2) and (10, oo), and falling on (2, 10). There is a local maximum at x = 2 and a local minimum at x = 10. The curve is concave down on (— oo, 4) and concave up on (4, oo). At x = 4 there is a point of inflection. 23. When y = x + sin x, then y' = 1 + cos x and y" = —sin x. The curve rises on (0, 2ir). At x = there is a local and absolute minimum and at x = 2tt there is a local and absolute maximum. The curve is concave down on (0, tt) and concave up on (7r, 27r). At x = 7T there is a point of inflection. y " Max 2ir - (2tt,2tt)% y = x + sin x / (it, tt) y ■n y^iun /Min I) IT 1lT 24. When y = x — sin x, then y' = 1 — cos x and y" = sin x. The curve rises on (0, 2ir). At x = there is a local and absolute minimum and at x = 2ir there is a local and absolute maximum. The curve is concave up on (0, tt) and concave down on (tt, 2ir). At x = 7r there is a point of inflection. A v-9/5 25 A 25. When y = x 1 / 5 , then y' = | x" 4 / 5 and y" The curve rises on (— oo, oo) and there are no extrema. The curve is concave up on (— oo, 0) and concave down on (0, oo). At x = there is a point of inflection. (0,0) Infl Copyright (c) 2006 Pearson Education 232 Chapter 4 Applications of Derivatives 26. When y = x 3 / 5 , then / = | x~ 2 / 5 and y" = - ^ x~ 7 / 5 . The curve rises on (— oo, oo) and there are no extrema. The curve is concave up on (— oo, 0) and concave down on (0, oo). At x = there is a point of inflection. 27. When y = x 2 / 5 , then y' = § x" 3 / 5 and y" = - ^ x~ 8 / 5 . The curve is rising on (0, oo) and falling on (— oo, 0). At x = there is a local and absolute minimum. There is no local or absolute maximum. The curve is concave down on (— oo, 0) and (0, oo). There are no points of inflection, but a cusp exists at x = 0. 28. When y = x 4 / 5 , then y' = f x" 1 / 5 and y" = - ^ x" 6 / 5 . The curve is rising on (0, oo) and falling on (— oo, 0). At x = there is a local and absolute minimum. There is no local or absolute maximum. The curve is concave down on (— oo, 0) and (0, oo). There are no points of inflection, but a cusp exists at x = 0. 29. When y = 2x - 3x 2 / 3 , then y' = 2 - 2X" 1 / 3 and y" = | x~ 4 / 3 . The curve is rising on (— oo, 0) and (1, oo), and falling on (0, 1). There is a local maximum at x = and a local minimum at x = 1. The curve is concave up on (— oo, 0) and (0, oo). There are no points of inflection, but a cusp exists at x = 0. Cusp, Loc max (0,0) 30. When y = 5x 2 / 5 - 2x, then y' = 2x~ 3 / 5 -2 = 2 (x^ 3 / 5 - l) and y" = — | x -8 / 5 . The curve is rising on (0, 1) and falling on (— oo, 0) and (1, oo). There is a local minimum at x = and a local maximum at x = 1. The curve is concave down on (— oo, 0) and (0, oo). There are no points of inflection, but a cusp exists at x = 0. (0,0) (1,3) y = 5x vi - 2x Copyright (c) 2006 Pearson Education Section 4.4 Concavity and Curve Sketching 233 31. Wheny = x 2 / 3 (§ 5 3 x) = | x 2/3 - x 5/3 , then ./ _ 5 Y -l/3 _ 5 Y 2/3 y — i A i A 3 | x - 1 /3(i _ x )and nX - 4 / 3 (l +2x). v" — - 5 y- 4 /3 _ 10 Y -l/3 y — 9 A 9 A The curve is rising on (0, 1) and falling on (— oo, 0) and (1, oo). There is a local minimum at x = and a local maximum at x = 1. The curve is concave up on (-co, — \) and concave down on (— i OJ and (0, oo). There is a point of inflection at x = — \ and a cusp at x = 0. 32. When y = x 2 / 3 (x - 5) = x 5 / 3 - 5x 2 / 3 , then y' = § x 2 / 3 h> x -i/3 = 5 X -1/3( X _ 2 ) a nd -1/3 -4/3 f x~ 4 / 3 (x + 1). The curve y - -g A -r -g- is rising on (— oo, 0) and (2, oo), and falling on (0, 2). There is a local minimum at x = 2 and a local maximum at x = 0. The curve is concave up on (—1, 0) and (0, oo), and concave down on (— oo, — 1). There is a point of inflection at x = — 1 and a cusp at x = 0. (*-5) 1-1,-6) (2.0,-4.76) 33. When y = x\/$ - x 2 = x (8 - x 2 ) 1/2 , then „2\l/2 )^ + (x)(i)(8-x 2 )- I/ V2x) x 2 ) 1/2 (8-2x 2 ) 2(2 - x)(2 + x) and 2^/2 + xj (2V2-X) 2\ i to „2\~2C -i)(8-x 2 ) 2 (-2x)(8-2x 2 ) + (8-x 2 ) 2 (-4x) 2x(x 2 -12) \/(8-x 2 ) 3 . The curve is rising on (—2, 2), and falling on ( — 2a/ 2, —21 and I 2, 2y 2) . There are local minima x = — 2 and x = 2 \J 2, and local maxima at x = — 2y 2 and x = 2. The curve is concave up on ( — 2y 2, 0] and concave down on 10,2 y 2 J . There is a point of inflection at x = 0. 34. When y = (2 - x 2 ) 3/2 , then y' = (|) (2 - x 2 ) 1/2 (-2x) = — 3xy 2 — x 2 = — 3x .2*1/2 2 + x j and v" : - ( -3) (2 - x 2 ) I/2 + (-3x) (I) (2 - x 2 )- 1/2 (-2x) -6(1-X)(1+X) rj,, ... ^^-^^^^^— . The curve is rising on v/2-x) (%/2 + x) — v 2, ) and falling on I 0, v 2 J . There is a local maximum at x = 0, and local minima at x = ± \J 2. The curve is concave down on (— 1, 1) and concave up on I — v 2, — 1 j and ( 1 , \J 2 j . There are points of inflection at x= ± 1. v Loc max •»\ <2,4) 4 / 3 - / 2 - / Loc max (-2^2, 0) 1 7(0,0) Infl 1 2 1 / 1 2 (-2^,0)' A Loc min 1 -3 - y = jcV8-JC 2 -4 _ (_2,_4)\K Loc mill (0,2/2") (-^",0) Copyright (c) 2006 Pearson Education 234 Chapter 4 Applications of Derivatives 35. Wheny ■ 3 then y' 2x(x-2)-(x 2 -3)(l) x - 2 ' " lk -" 1 3 (x-2) 2 (x-3)(x-l) A (x - 2)2 an<1 (2x - 4)(x - 2) 2 - (x 2 - 4x + 3)2 (x - 2) _ J (x - 2) 4 ~ (x - 2) 3 • The curve is rising on (—00, 1) and (3, 00), and falling on (1, 2) and (2, 3). There is a local maximum at x = 1 and a local minimum at x = 3. The curve is concave down on (—00, 2) and concave up on (2, 00). There are no points of inflection because x = 2 is not in the domain. 36. Wheny = 3^, then y _ 3x 2 (x 2 + l) / _ 3x 2 (3x 2 +l)-x 3 (6x) (3x2 + 1) 2 (3x2 + 1)2 and „ _ (12x 3 + 6x)(3x 2 + l) -2(3x 2 + l)(6x)(3x 4 + 3x 2 ) y ~~ (3x2 + 1) 4 6x(l-x)(l+x) The curve is rising on (—00, 00) so (3x2 + l) J there are no local extrema. The curve is concave up on (—00, —1) and (0, 1), and concave down on (— 1, 0) and (1, 00). There are points of inflection at x = — 1, x = 0, and x = 1 . 3x* + l 37. Wheny = |x 2 - 1| -- 2x, Ixl > 1 !. I x l > 1 u 2 ~ , , then <-{ and y" The x 2 , |x| < 1 2, |x| > 1 2x, |x| < 1 """ J {-2, |x| < 1 curve rises on (—1, 0) and (1, 00) and falls on (—00, —1) and (0, 1). There is a local maximum at x = and local minima at x = ±1. The curve is concave up on (—00, —1) and (1, 00), and concave down on (— 1, 1). There are no points of inflection because y is not differentiable at x = ±1 (so there is no tangent line at those points). -2 (-1, 0) Loc min r 38. Wheny 2x1 2x - 2x, x < x 2 , < x < 2 then ^ x 2 - 2x, x > 2 ti x<0 < x< 2 . 2, x > 2 f 2x - 2, x < y' = < 2 - 2x, < x < 2 , and y" ^ 2x - 2, x > 2 ( The curve is rising on (0, 1) and (2, 00), and falling on (—00, 0) and (1, 2). There is a local maximum at x = 1 and local minima at x = and x = 2. The curve is concave up on (—00, 0) and (2, 00), and concave down on (0, 2). There are no points of inflection because y is not differentiable at x = and x = 2 (so there is no tangent at those points). Copyright (c) 2006 Pearson Education Section 4.4 Concavity and Curve Sketching 235 39. Wheny I, 2 V^ Since lim y -oo and lim y x^0+ 0. There is a local minimum at x oo there is a cusp at x = 0. There is a local minimum at x = 0, but no local maximum. The curve is concave down on (— oo, 0) and (0, oo). There are no points of inflection. 40. Wheny=y^4T= J V^Z> x > 4 [ a/4-x, x<4 then H5 , x > 4 , x < 4 and y" _ (x _ 4) -3/2 -(4-x)- x > 4 x < 4 Since lim y' = — oo and lim y' = oo there is a cusp x -» 4 x -> 4+ at x = 4. There is a local minimum at x = 4, but no local maximum. The curve is concave down on (— oo, 4) and (4, oo). There are no points of inflection. 1 1 2 - ^_^- N.V^(ao) i ] r .^ i i i i ^ -4 -3 -2 - i \ 1 2 3 4 Cusp Loc min (4,0) 41. y' = 2 + x - x 2 = (1 + x)(2 - x), y' = | +++ | — -1 2 =>■ rising on (—1, 2), falling on (— oo, —1) and (2, oo) =>■ there is a local maximum at x = 2 and a local minimum atx= -l;y" = 1 - 2x, y" = +++ | 1/2 =£> concave up on (— oo, |) , concave down on Q, oo) =>• a point of inflection at x = \ 42. y' = x 2 - x - 6 = (x - 3)(x + 2), y' = +++ | | ++- -2 3 => rising on (— oo, —2) and (3, oo), falling on (—2, 3) =4> there is a local maximum at x = —2 and a local minimum at x = 3; y" = 2x — 1, y" = | +++ 1/2 =4> concave up on (|, oo), concave down on (— oo, |) => a point of inflection at x = \ x — 2 43. y' = x(x - 3) 2 , y' = | +++ | +++ =>• rising on 3 (0, oo), falling on (—oo, 0) => no local maximum, but there is a local minimum at x = 0; y" = (x — 3) 2 + x(2)(x — 3) = 3(x - 3)(x - 1), y" = +++ | | +++ => concave 1 3 up on (— oo, 1) and (3, oo), concave down on (1, 3) =>• points of inflection at x = 1 and x = 3 Cop # (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 236 Chapter 4 Applications of Derivatives 44. y' = x 2 (2 - x), y' = +++ | +++ | =4> rising on 2 (— oo, 2), falling on (2, oo) => there is a local maximum at x = 2, but no local minimum; y" = 2x(2 — x) + x 2 (— 1) = x(4 — 3x), y" = | +++ | => concave up 4/3 on (0, |) , concave down on (— oo, 0) and (|, oo) of inflection at x = and x = i points 45. y' = x (x 2 - 12) = x (x - 2^) (x + 2^3) , y' = - -2JZ -+|— I +- 2^3 rising on 2 a/3, o) and (2^, 00) , falling on (-00, -2^/t\ and I 0, 2 v 3 ) =>■ a local maximum at x = 0, local minima at x = ± 2v/3 ; y" = (1) (x 2 - 12) + (x)(2x) = 3(x - 2)(x + 2), y" = +++ I I +++ -2 2 =$■ concave up on (—00, —2) and (2, 00), concave down on (—2,2) =^ points of inflection at x = ±2 46. y' = (x - l) 2 (2x + 3), y' = | +++ | +++ -3/2 1 =>- rising on (— |, 00) , falling on (—00, — |) =4> no local maximum, a local minimum at x = — | ; y" = 2(x - l)(2x + 3) + (x - 1) 2 (2) = 2(x - l)(3x + 2), +- -2/3 1 -+ concave up on ^ — 00, — |) and (1, 00), concave down on (— |, l) => points of inflection at x = — I and x = 1 x-1 'X--2/3 X--3/2 47. y' = (8x - 5x 2 ) (4 - x) 2 = x(8 - 5x)(4 - x) 2 , y' = | +++ | | =4> rising on (0, a local maximum at -I+++I — I- 8/5 4 falling on (—00, 0) and (I, 00) x = I , a local minimum at x = 0; y" = (8 - 10x)(4 - x) 2 + (8x - 5x 2 ) (2)(4 - x)(-l) = 4(4 - x) (5x 2 - 16x + 8) , y" = +++ | +++ | =► concave up 8-2^/6 8+2^/6 4 5 5 on I —00, ~g J and I + ^ , 4 J , concave down on -\* x 2 ^ and (4 j00 ) ^ points of inflection at and x = 4 5 ' 5 8±2\/6 Copffigl (c| 1 Pearson Education, to, publishing as Pearson Addison-Wesle Section 4.4 Concavity and Curve Sketching 237 y' = (x 2 - 2x) (x - 5) 2 = x(x - 2)(x - 5) 2 , y' = +++ | | +++ | +++ => rising on (-00, 0) 2 5 and (2, 00), falling on (0, 2) =4> a local maximum at x = 0, a local minimum at x = 2; y" = (2x - 2)(x - 5) 2 + (x 2 - 2x) (2)(x - 5) = 2(x-5)(2x 2 -8x + 5), -+- 4-V6 4-j-yjj 2 -+ =>- concave up ( — ^—, + 2 J and (5, 00), concave down on 4-^6 ) and I + ,/ , 5 ) => points of inflection at x = i±y^ and x = 5 x »0 x«2 x = 5 7T 7T \ 2> 2/ y = sec J x, y = ( +++ ) =>- rising on -tt/2 tt/2 never falling =£• no local extrema; y" = 2(sec x)(sec x)(tan x) = 2 (sec 2 x) (tan x), y" = ( I +++ ) => concave up on (0, | -tt/2 tt/2 concave down on (— § , 0) , is a opoint of inflection 50. y' = tan x, y' = ( | +++ ) =4> rising on (0, §) , -tt/2 tt/2 falling on (— |, 0) =>- no local maximum, a local minimum at x = 0; y" = sec 2 x, y" = ( +++ ) =4> concave up -tt/2 tt/2 on (— |, I) => no points of inflection 51. y' = cot f , y' = ( +++ I ) => rising on (0,tt), T 2tt falling on (7r, 27r) =4- a local maximum at = 7r, no local minimum; y" = — \ esc 2 f , y" = ( ) => never 2tt concave up, concave down on (0, 2ir) => no points of inflection 52. y' = esc 2 I , y 1 = ( +++ ) =4> rising on (0, 2*7r), never 2tt falling =S> no local extrema; y" = 2(csc" v -csc 5 J (cot ~ 2 f] 2; f 2tt concave up on (7r, 27t), concave down on (0, 7r) a point of inflection at 9 = n Copffigl (c| 1 Pearson Education, Ire, publishing as Pearson Addison-Wesle 238 Chapter 4 Applications of Derivatives 53. y' = tan 2 6-1 = (tan 9 - l)(tan 9 + 1), y' = ( +++| | +++) =4> rising on -tt/2 -7r/4 tt/4 tt/2 (-§,-£) and (f,§), falling on (-£,f) =^> a local maximum at y" = 2 tan 6> sec 2 9, y" = ( a local minimum at 9 — I+++) -tt/2 tt/2 concave up on (0, |) , concave down on (— |, Oj a point of inflection at 9 = 54. y' = 1 - cot 2 6> = (1 - cot 6>)(1 + cot 0), y' = ( I +++I ) => rising on (|, 3 f) tt/4 3tt/4 w falling on (0, |) and (x> 7r ) =^ a l° ca l maximum at 6> = ^, a local minimum at 9 = | ; y" = -2(cot 61) (-esc 2 9), y" = ( +++ | ) tt/2 7r =^> concave up on (0, |) , concave down on (|, 7r) =4> a point of inflection at # = | e=3it/4 rising on 55. y' = cost, y' = [+++ | | +++] tt/2 3tt/2 2tt (0, §) and (^f , 27r) , falling on (§, ^) => local maxima at t = | and t = 27T, local minima at t = and t = 4r ; y" = -sint, y" = [ | - 1" 2tt => concave up on (tt, 2ir), concave down on (0, 7r) => a point of inflection at t = 7r 56. y' = sin t, y' = [ +++ | ] =4> rising on (0, it), tt 2tt falling on (tt, 2ir) =$■ a local maximum at t = n, local minima at t = and t = 27r; y" = cos t, y" = [+++| | +++] ^ concave up on (0, |) tt/2 3tt/2 2tt and , 27T] , concave down on 2' 2 points of inflection at t = | and t = y t-3it/2 57. y' = (x+l) -2/3 - + +) (+- -1 rising on (— oo, oo), never falling => no local extrema; y" = - | ( X + l)-5/3 ; y " = +++ ) ( -1 =>- concave up on (— oo, —1), concave down on (—1, oo) =^ a point of inflection and vertical tangent at x = — 1 Infl Veit tan Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 4.4 Concavity and Curve Sketching 239 58. y' = (x - 2)- 1 / 3 , y' = )(• rising on (2, oo), falling on (— oo, 2) => no local maximum, but a local minimum at x = 2; y" = — | (x — 2)~ 4 / 3 , y" = )( => concave down on (— oo,2) and 2 (2, oo) => no points of inflection, but there is a cusp at x = 2 59. y' = x~ 2 / 3 (x - 1), y' = )( | +++ =>• rising on 1 (1, oo), falling on (—oo, 1) =4> no local maximum, but a local minimum at x = 1; y" = = x -2 / 3 + \ x~ 5 / 3 = \ x- 5 / 3 (x + 2), y" = +++ | )( +++ -2 =£> concave up on (—oo, —2) and (0, oo), concave down on (—2,0) =^ points of inflection at x = —2 and x = 0, and a vertical tangent at x = 60. y' = x" 4 / 5 (x + 1), y' = | +++)(+++ =>• rising on -1 (—1,0) and (0, oo), falling on (—oo, —1) => no local maximum, but a local minimum at x = — 1 ; y" = I x-1/5 _ 4 x -9/5 = 1 x -9/5 (x _ 4)> y" = +++ )( | +++ => concave up on (—00, 0) and 4 (4, 00), concave down on (0, 4) =>■ points of inflection at x = and x = 4, and a vertical tangent at x = 61. y' -2x, x < , 2x, x > ' y rising on (—00, 00) => no local extrema -"={t x<0 x>0 ' y" = )( +++ => concave up on (0, 00), concave down on (—00,0) => a point of inflection at x = 62. y' -x 2 , x < , x 2 , x > ' y rising on (0, 00), falling on (—00, 0) => no local maximum, but a , • • ^ 11 \ ~2x, x < local minimum at x = 0; y = < „ „, J \ 2x, x > +- concave up on (—00, 00) no point of inflection Cop # (c) 1 Pearson Education, to, publishing as Pearson Addison-Wesle 240 Chapter 4 Applications of Derivatives 63. The graph of y = f "(x) =>- the graph of y up on (0, oo), concave down on (—oo,0) - inflection at x = 0; the graph of y = f '(x) =>■ y 1 = +++ | | +++ => the graph y both a local maximum and a local minimum f(x) is concave a point of f(x) has 64. The graph of y = f"(x) =>• y" = +++ | =* the graph of y = f(x) has a point of inflection, the graph of y = f'( x ) => y' = | +++ | =4> the graph of y = f(x) has both a local maximum and a local minimum 65. The graph of y = f"(x) =► y" = | +++ | =4> the graph of y = f(x) has two points of inflection, the graph of y = f'(x) => y' = | +++ =>■ the graph of y = f(x) has a local minimum y > . p ^— y N^Infl In il*C nX Loc N. rain y" 66. The graph of y = f"(x) ^ y" = +++ | ^> the graph of y = f(x) has a point of inflection; the graph of y = f'( x ) =>■ y' = I +++ I => the graph of y = f(x) has both a local maximum and a local minimum 67. Point y' y" P - + Q + R + - S - T - - Copyright (c) 2006 Pearson Education Section 4.4 Concavity and Curve Sketching 241 68. 69. 7 - (6,7) \ 4 1 JXV (4,4) 2 4 6 70. 71. Graphs printed in color can shift during a press run, so your values may differ somewhat from those given here. (a) The body is moving away from the origin when |displacement| is increasing as t increases, < t < 2 and 6 < t < 9.5; the body is moving toward the origin when |displacementj is decreasing as t increases, 2 < t < 6 and 9.5 < t < 15 (b) The velocity will be zero when the slope of the tangent line for y = s(t) is horizontal. The velocity is zero when t is approximately 2, 6, or 9.5 sec. (c) The acceleration will be zero at those values of t where the curve y = s(t) has points of inflection. The acceleration is zero when t is approximately 4, 7.5, or 12.5 sec. (d) The acceleration is positive when the concavity is up, 4 < t < 7.5 and 12.5 < t < 15; the acceleration is negative when the concavity is down, < t < 4 and 7.5 < t < 12.5 72. (a) The body is moving away from the origin when |displacement| is increasing as t increases, 1.5 < t < 4, 10 < t < 12 and 13.5 < t < 16; the body is moving toward the origin when | displacement! i s decreasing as t increases, < t < 1.5, 4 < t < 10 and 12 < t < 13.5 (b) The velocity will be zero when the slope of the tangent line for y = s(t) is horizontal. The velocity is zero when t is approximately 0, 4, 12 or 16 sec. (c) The acceleration will be zero at those values of t where the curve y = s(t) has points of inflection. The acceleration is zero when t is approximately 1.5, 6, 8, 10.5, or 13.5 sec. (d) The acceleration is positive when the concavity is up, < t < 1.5, 6 < t < 8 and 10 < t < 13.5, the acceleration is negative when the concavity is down, 1.5 < t < 6, 8 < t < 10 and 13.5 < t < 16. 73. The marginal cost is ^ which changes from decreasing to increasing when its derivative *p is zero. This is a point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units. dy 74. The marginal revenue is -^ and it is increasing when its derivative -A is positive => the curve is concave up d 2 y < t < 2 and 5 < t < 9; marginal revenue is decreasing when jp < 2 < t < 5 and 9 < t < 12. the curve is concave down 75. When y' = (x - l) 2 (x - 2), then y" = 2(x - l)(x - 2) + (x - l) 2 . The curve falls on (-oo, 2) and rises on (2, oo). At x = 2 there is a local minimum. There is no local maximum. The curve is concave upward on (— oo, 1) and Copyright (c) 1 ton E Won, Inc., puiing as Pearson Addison-Wesle 242 Chapter 4 Applications of Derivatives (|, oo) , and concave downward on (l, |) . At x = 1 or x = | there are inflection points. 76. When y' = (x - l) 2 (x - 2)(x - 4), then y" = 2(x - l)(x - 2)(x - 4) + (x - l) 2 (x - 4) + (x - l) 2 (x - 2) = (x - 1) [2 (x 2 - 6x + 8) + (x 2 - 5x + 4) + (x 2 - 3x + 2)] = 2(x - 1) (2x 2 - lOx + 11). The curve rises on (— oo, 2) and (4, oo) and falls on (2, 4). At x = 2 there is a local maximum and at x = 4 a local minimum. The curve is concave downward on (— oo, 1) and I ~£ , ~y 1 and concave upward on ( 1, ~£ I and I ^ , oo ) . At x = 1, — ^— and ~y there are inflection points. 77. The graph must be concave down for x > because f"(x) = - 4 < 0. 78. The second derivative, being continuous and never zero, cannot change sign. Therefore the graph will always be concave up or concave down so it will have no inflection points and no cusps or corners. 79. The curve will have a point of inflection at x = 1 if 1 is a solution of y" = 0; y = x 3 + bx 2 + ex + d 3x 2 + 2bx + c 6x + 2band6(l) + 2b = 80. (a) True. If f(x) is a polynomial of even degree then f ' is of odd degree. Every polynomial of odd degree has at least one real root =>■ f '(x) = for some x = r => f has a horizontal tangent at x = r. (b) False. For example, f(x) = x — 1 is a polynomial of odd degree but f '(x) = 1 is never 0. As another example, y = I x 3 + x 2 + x is a polynomial of odd degree, but y' = x 2 + 2x + 1 = (x + l) 2 > for all x. 81. (a) f(x) = ax 2 + bx + c = a (x 2 + \ x) + c = a (x 2 + \ x + ^) - g + c = a (x + ^f whose vertex is at x = — £■ =>- the coordinates of the vertex are ( — j- , — b 4a 4ac ) b 2 -4ac 4a a parabola (b) The second derivative, f "(x) = 2a, describes concavity when a < the parabola is concave down. when a > the parabola is concave up and 82. No, f"(x) could be decreasing to zero at x = c and then increase again so it would be concave up on every interval even though f"(x) = 0. For example f(x) = x 4 is always concave up even though f"(0) = 0. 83. A quadratic curve never has an inflection point. If y = ax 2 + bx + c where a^0, then y' = 2ax + b and y" = 2a. Since 2a is a constant, it is not possible for y" to change signs. 84. A cubic curve always has exactly one inflection point. If y = ax 3 + bx 2 + ex + d where a ^ 0, then y' = 3ax 2 + 2bx + c and y" = 6ax + 2b. Since ^ is a solution of y" = 0, we have that y" changes its sign at x = — j- and y' exists everywhere (so there is a tangent at x = — £-). Thus the curve has an inflection point at x = — ^- . There are no other inflection points because y" changes sign only at this zero. Copyright (c) 2006 Pearson Education Section 4.4 Concavity and Curve Sketching 243 85. If y = x 5 - 5x 4 - 240, then y' = 5x 3 (x - 4) and y" = 20x 2 (x — 3). The zeros of y' are extrema, and there is a point of inflection at x = 3. v 200 - y i = 20x 2 (j:-3)/ / -200 ^ -X^3 A /=5.t 3 (*-4) 5x 4 - 240 s v_y 5 -400 y- =* 5 - 86. If y = x 3 - 12x 2 , then y' = 3x(x - 8) and y" = 6(x — 4). The zeros of y' and y" are extrema and points of inflection, respectively. : N. 50 y' -6(x-41^ "iU^J ^2*^ 4 i^S /^50 -100 y' » 3x(x - 8) -150 -200 \y-x 3 -12x 2 -250 87. If y = f x 5 + 16x 2 - 25, then y' = 4x (x 3 + 8) and y" = 16 (x 3 + 2) . The zeros of y' and y" are extrema and points of inflection, respectively. , y \ 100 \ 50 -3/ \y_>^ _^S' 2 3 '" / / ~ 50 - > = ijc 5 + 16x 2 -25 ' / -100 "=16(x 3 + 2) 88. Ify r - f -4x 2 + 12x + 20, then y' = x 3 - x 2 - 8x + 12 = (x + 3)(x - 2) 2 . So y has a local minimum at x = — 3 as its only extreme value. Also y" = 3x 2 - 2x - 8 = (3x + 4)(x - 2) and there are inflection points at both zeros, — | and 2, of y". y"-(3x + 4)(x-2) y-T-, -4X..+ 12X + 20 21. The graph of f falls where f ' < 0, rises where f ' > 0, and has horizontal tangents where f = 0. It has local minima at points where f ' changes from negative to positive and local maxima where f ' changes from positive to negative. The graph of f is concave down where f " < and concave up where f " > 0. It has an inflection point each time f "changes sign, provided a tangent line exists there. y f\ r\ rl / /f f / Copyright (c) 2006 Pearson Education 244 Chapter 4 Applications of Derivatives 90. The graph f is concave down where f " < 0, and concave up where f " > 0. It has an inflection point each time f " changes sign, provided a tangent line exists there. 91. (a) It appears to control the number and magnitude of the local extrema. If k < 0, there is a local maximum to the left of the origin and a local minimum to the right. The larger the magnitude of k (k < 0), the greater the magnitude of the extrema. If k > 0, the graph has only positive slopes and lies entirely in the first and third quadrants with no local extrema. The graph becomes increasingly steep and straight as k — ► oo. (b) f '(x) = 3x 2 + k =>■ the discriminant 2 negative for k > 0; f ' has two zeros x = when k > 0; the sign of k controls the number of local extrema (c) As k — > oo, f'(x) — > oo and the graph becomes increasingly steep and straight. As k — > — oo, the crest of the graph (local maximum) in the second quadrant becomes increasingly high and the trough (local minimum) in the fourth quadrant becomes increasingly deep. - 4(3)(k) = — 12k is positive for k < 0, zero for k = 0, and ± J — | when k < 0, one zero x = when k = and no real zeros y , 200 N. \100 J/ ~y v^ 5 " ° ^?\— 2~~" k/ / / / -100 \v_ -200 92. (a) It appears to control the concavity and the number of local extrema. k=-10 Copyright (c) 2006 Pearson Education Section 4.5 Applied Optimization Problems 245 (b) f(x) = x 4 + kx 3 + 6x 2 => f (x) = 4x 3 + 3kx 2 + 12x => f"(x) = 12x 2 + 6kx +12 => the discriminant is 36k 2 - 4(12)(12) = 36(k + 4)(k - 4), so the sign line of the discriminant is +++ | | +++ =>• the -4 4 discriminant is positive when |k| > 4, zero when k = ± 4, and negative when |k| < 4; f"(x) = has two zeros when |k| > 4, one zero when k = ± 4, and no real zeros for |k| < 4; the value of k controls the number of possible points of inflection. 93. (a) If y = x 2 / 3 (x 2 - 2) , then y' = f x" 1 / 3 (2x 2 - 1) and y" = | x~ 4 ' 3 (10x 2 + 1) . The curve rises on (- 4j , OJ and ( X , ooj and falls on (-co, - -^j and The curve is concave up on (— oo, 0) and (0, oo ). (b) A cusp since lim y' = oo and lim y' x -» x -» 0+ y=X 1 '\x 1 -2) 94. (a) If y = 9x 2 / 3 (x - 1), then y' „ _ 10 (x+l) 15 (x- xV3 and X 1 3 The curve rises on (— oo, 0) and I, oo) and falls on (0, 1) . The curve is concave down on (0,oo). -oo, - i) and concave up on (— 1, 0) and (b) A cusp since lim y' x — > oo and lim y' x^0+ 95. Yes: y = x + 3 sin 2x =4> y' = 2x + 6 cos 2x. The graph of y' is zero near —3 and this indicates a horizontal tangent near x = —3. 2x + 6 cos 2x 4.5 APPLIED OPTIMIZATION PROBLEMS 32 _ 2 (I 2 -16) 1. Let £ and w represent the length and width of the rectangle, respectively. With an area of 16 in. 2 , we have that (f)(w) = 16 => w = let 1 => the perimeter is P = U + 2w = 2£ + 12tr l and P'(£) = 2 Solving P'(£) = =^ 2(f + 4K£-4) = Q ^ £ = _4 5 4 since £ > o for the length of a rectangle, I must be 4 and w = 4 =$■ the perimeter is 16 in., a minimum since P"(£) = || > 0. 2. Let x represent the length of the rectangle in meters (0 < x < 4) Then the width is 4 — x and the area is A(x) = x(4 - x) = 4x - x 2 . Since A'(x) = 4 - 2x, the critical point occurs at x = 2. Since, A'(x) > for < x < 2 and A'(x) < for 2 < x < 4, this critical point corresponds to the maximum area. The rectangle with the largest area measures 2 m by 4 — 2 = 2 m, so it is a square. Copyright (c) 2006 Pearson Education 246 Chapter 4 Applications of Derivatives Graphical Support: A(x) (a) The line containing point P also contains the points (0, 1) and (1, 0) => the line containing P is y = 1 — x => a general point on that line is (x, 1 — x). (b) The area A(x) = 2x(l — x), where < x < 1. (c) When A(x) = 2x - 2x 2 , then A'(x) = => 2-4x = => x = ± . Since A(0) = and A(l) = 0, we conclude that A (i) = | sq units is the largest area. The dimensions are 1 unit by | unit. The area of the rectangle is A = 2xy = 2x ( 1 2 — x 2 ) , where < x < ^f\2 . Solving A'(x) = =4> 24 - 6x 2 = =4> x = —2 or 2. Now —2 is not in the domain, and since A(0) = and A (\f\2\ = 0, we conclude that A(2) = 32 square units is the maximum area. The dimensions are 4 units by 8 units. /\ a \ The volume of the box is V(x) = x(15 — 2x)(8 — 2x) = 120x - 46x 2 + 4x 3 , where < x < 4. Solving V'(x) =>■ 120 - 92x + 12x 2 = 4(6 - x)(5 - 3x) = => x = or 6, but 6 is not in the domain. Since V(0) = V(4) = 0, V (|) = 2 |ip ~ 91 in 3 must be the maximum volume of the box with dimensions yX fx ? inches. 15-2X 1 1 6. The area of the triangle is A = | ba = | y 400 — b 2 , where < b < 20. Then ^ = ± V '400 - b 2 , h " - db 2 v 2\/40()-b2 = 2 P°' h \ = =^> the interior critical point is b = 10\/2. V400 - b 2 f V When b = or 20, the area is zero => A I 10 v 2 J is the maximum area. When a 2 + b 2 = 400 and b = 10y 2, the value of a is also 10 y 2 a = b. the maximum area occurs when (0.b) . \20 («.0) 7. The area is A(x) = x(800 - 2x), where < x < 400. Solving A'(x) = 800 - 4x = => x = 200. With A(0) = A(400) = 0, the maximum area is A(200) = 80,000 m 2 . The dimensions are 200 m by 400 m. rlvsr 800-2X Copyright (c) 2006 Pearson Education Section 4.5 Applied Optimization Problems 247 The area is 2xy = 216 needed is P = 4x + 3y g=4-ff=0=>x => y = = 4x4 2 S j — . The amount of fence X 324x _1 , where < x; = => the critical points are and ± 9, but and —9 are not in the domain. Then P"(9) > =4> at x = 9 there is a minimum =>- the dimensions of the outer rectangle are 18 m by 12 m => 72 meters of fence will be needed. (a) We minimize the weight = tS where S is the surface area, and t is the thickness of the steel walls of the tank. The its di ?200). Treating the surace area is S = x 2 + 4xy where x is the length of a side of the square base of the tank, and y is its depth. The volume of the tank must be 500ft 3 => y = ™ . Therefore, the weight of the tank is w(x) = t(x 2 thickness as a constant gives w'(x) = t(2x — ^2) for x.0. The critical value is at x = 10. Since w"(10) = t(2 + ^gr) > 0, there is a minimum at x = 10. Therefore, the optimum dimensions of the tank are 10 ft on the base edges and 5 ft deep, (b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of the steel walls would likely be determined by other considerations such as structural requirements. 10. (a) The volume of the tank being 1125 ft 3 , we have that yx 2 = 1125 c(x) = 5x 2 + 30x(±±P), where < x. Then c'(x) = lOx X 2 ii|2 . The cost of building the tank is ■ the critical points are and 15, but is not in the domain. Thus, c"(15) > => at x = 15 we have a minimum. The values of x = 15 ft and y = 5 f t will minimize the cost, (b) The cost function c = 5(x 2 + 4xy) + lOxy, can be separated into two items: (1) the cost of the materials and labor to fabricate the tank, and (2) the cost for the excavation. Since the area of the sides and bottom of the tanks is (x 2 + 4xy), it can be deduced that the unit cost to fabricate the tanks is $5/ft 2 . Normally, excavation costs are per unit volume of excavated material. Consequently, the total excavation cost can be taken as lOxy = (— ) (x 2 y). This suggests that the S10/ft 2 unit cost of excavation is ^—^ — where x is the length of a side of the square base of the tank in feet. For the least " 18 The total cost of the least expensive tank is expensive tank, the unit cost for the excavation is 1 / ft t - ■■-—■ $3375, which is the sum of $2625 for fabrication and $750 for the excavation. sn.fir ft3 11. The area of the printing is (y — 4)(x — 8) = 50. Consequently, y = (j^g) + 4. The area of the paper is A(x) = x (^ + 4) , where 8 < x. Then A'(x) 50 + 4 -x [o^Wj 4(x-8) 2 -400 vx-8 ' 7 \(x-syj (x-8) 2 => the critical points are —2 and 18, but —2 is not in the domain. Thus A"(18) > => atx = 18 we have a minimum. Therefore the dimensions 18 by 9 inches minimize the amount of paper. 4 2 2 oo i X y-4 4 12. The volume of the cone is V = h 7rr 2 h, where r = x = i/9 — y 2 and h = y + 3 (from the figure in the text). Thus, V(y) = f (9 - y 2 ) (y + 3) = § (27 + 9y - 3y 2 - y 3 ) => V'(y) = § (9 - 6y - 3y 2 ) = tt(1 - y)(3 + y). The critical points are —3 and 1, but —3 is not in the domain. Thus V"(l) = | (—6 — 6(1)) < =£- at y = 1 we have a maximum volume of V(l) = | (8)(4) = =y£ cubic units. Copyright (c) 2006 Pearson Education 248 Chapter 4 Applications of Derivatives 13. The area of the triangle is A(0) ab sin 6 Solving A'(0) = ab cos where < < ir. = f. Since A"(0) A" (|) < 0, there is a maximum at 14. A volume V = 7rTh = 1000 h 1000 The amount of material is the surface area given by the sides and bottom of the can dS _ dr are and 2000 _ r 2 10 3/3 = 27rrh + 7rr 2 2ttt= 2000 TIT 3 - 1000 7rr 2 , < r. Then = 0. The critical points but is not in the domain. Since d_s _ 4goo _|_ 2tj- > o, we have a minimum surface area when r = 4t= cm an d h = i^r = 4t= cm - Comparing this result to the result found in Example 2, if we include both ends of the can, then we have a minimum surface area when the can is shorter-specifically, when the height of the can is the same as its diameter. 15. With a volume of 1000 cm and V = 7rr 2 h, then h = A = 8r 2 + 27rrh = 8r 2 + 2S2Q _ Then A ' (r) = 16r The amount of aluminum used per can is 2000 r 2 8r 3 -1000 =>■ the critical points are and 5, 40 but r = results in no can. Since A (r) = 16 + ^tt > we have a minimum atr = 5 =>■ h = — and h:r = 8 :tt. 16. (a) The base measures 10 — 2x in. by 15-2x in., so the volume formula is V(x) x(10-2x)(15-2x) 2 2x 3 - 25x 2 + 75x. (b) We require x > 0, 2x < 10, and 2x < 15. Combining these requirements, the domain is the interval (0, 5). V 80- 60 - 40 20 -20 (c) The maximum volume is approximately 66.02 in. 3 when x w 1.96 in. (1.9618739,66.019118) (d) V'(x) = 6x 2 — 50x + 75. The critical point occurs when V'(x) = 0, at x 50± v /(-50) 2 -4(6)(75) _ 50 ±V700 2(6) — 12 = ' J - v , that is, x w 1.96 or x w 6.37. We discard the larger value because it is not in the domain. Since V"(x) = 12x — 50, which is negative when x w 1.96 , the critical point corresponds to the maximum volume. The 25-5^ maximum volume occurs when x 6 1.96, which comfimrs the result in (c). 17. (a) The" sides" of the suitcase will measure 24 — 2x in. by 18 — 2x in. and will be 2x in. apart, so the volume formula is V(x) = 2x(24 - 2x)(18 - 2x) = 8x 3 - 168x 2 + 862x. Copyright (c) 2006 Pearson Education Dt Section 4.5 Applied Optimization Problems 249 (b) We require x > 0, 2x < 18, and 2x < 24. Combining these requirements, the domain is the interval (0, 9). -400 (c) The maximum volume is approximately 1309.95 in. 3 when x « 3.39 in. 1600 1 (3.3944503,1309.9547) 1200 -400 (d) V'(x) = 24x 2 - 336x + 864 = 24(x 2 - 14x + 36). The critical point is at x 14±y / (-14) 2 -4(l)(36) i 4±> /52 2(1) 2 = 7 ± y 13, that is, x w 3.39 or x « 10.61. We discard the larger value because it is not in the domain. Since V"(x) = 24(2x — 14) which is negative when x w 3.39, the critical point corresponds to the maximum volume. The maximum value occurs at x = 7 — y 13 w 3.39, which confirms the results in (c). (e) 8x 3 - 168x 2 + 862x = 1120 => 8(x 3 - 21x 2 + 108x - 140) = => 8(x - 2)(x - 5)(x - 14) = 0. Since 14 is not in the fomain, the possible values of x are x = 2 in. or x = 5 in. (f) The dimensions of the resulting box are 2x in., (24 — 2x) in., and (18 — 2x). Each of these measurements must be positive, so that gives the domain of (0, 9). 18. If the upper right vertex of the rectangle is located at (x, 4 cos 0.5 x) for < x < it, then the rectangle has width 2x and height 4 cos 0.5x, so the area is A(x) = 8x cos 0.5x. Solving A'(x) = graphically for < x < n, we find that x w 2.214. Evaluating 2x and 4 cos 0.5x for x ss 2.214, the dimensions of the rectangle are approximately 4.43 (width) by 1.79 (height), and the maximum area is approximately 7.923. 19. Let the radius of the cylinder be r cm, < r < 10. Then the height is 2y 100 — r 2 and the volume is V(r) = 271-rVlOO-r 2 cm 3 . Then, V'(r) = 2m 2 x/100-r 2 (-2r) + (27^100 - r 2 ) (2r) -2m 3 + 47ri-(100-r 2 ) \/l00-r 2 27tt(200 - 3f 2 ) VlOO-r 2 The critical point for < r < 10 occurs at r : 10 . Since V'(r) > for < r < 10a/ I and V'(r) < for 10W | < r < 10, the critical point corresponds to the maximum volume. The dimensions are r = 10 .16 cm and h 20 11.55 cm, and the volume is 4000?r 3\/3 2418.40 cm 3 20. (a) From the diagram we have 4x + I = 108 and V = x 2 L The volume of the box is V(x) = x 2 (108 — 4x), where < x < 27. Then V'(x) = 216x - 12x 2 = 12x(18 - x) = =>• the critical points are and 18, but x = results in no box. Since V"(x) = 216 - 24x < at x = 18 we have a maximum. The dimensions of the box are 18 x 18 x 36 in. Copyright (c) 2006 Pearson Education 250 Chapter 4 Applications of Derivatives (b) In terms of length, V(f) = x 2 £ ;^) 2 £. The graph indicates that the maximum volume occurs near I = 36, which is consistent with the result of part (a). 21. (a) From the diagram we have 3h + 2w V = h 2 w => V(h) = h 2 (54 - \ h) 108 and 54h 2 - | h 3 . Then V'(h) = 108h - § h 2 = f h(24 - h) = =>- h = or h = 24, but h = results in no box. Since V"(h) = 108 - 9h < at h = 24, we have a maximum volume at h = 24 and w = 54 - \ h = 18. (b) 10 15 20 25 30 35 22. From the diagram the perimeter is P = 2r + 2h + 7rr, where r is the radius of the semicircle and h is the height of the rectangle. The amount of light transmitted proportional to A = 2rh + \ tit 2 = r(P - 2r - ?rr) + \m 2 rP 4 2r 2 - \m 2 Then ^ dr P - 4r - i Trr = 2P 8 + 3- 2r _ 2h 4P 8 + 3- 2ttP 8 + 3tt (4 + tt)P 8 + 3- Therefore, £ ' h most light since ^— gives the proportions that admit the 4 di-2 |tt<0. 2r 23. j3 — M , where h is the height of the cylinder and r is the radius The fixed volume is V = 7rr 2 h + § 7rr 3 =>■ h of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the surface area of the hemisphere. Thus, we minimize C = 27rrh + 47rr 2 = 27rr (^ — If) + 47rr 2 = 2 j- => V | Trr 2 . Then f -- 4VV3 f + 1 i*r: fTrr 3 1^) . From the volume equation, h = ^ — y 2-3i/3.yi/3 _ 3i/3. 2 .4.yi/3 _ 2-3i/3. v i/3 ^1/3.32/3 3-2-7rV3 3-2-tt1/3 dimensions do minimize the cost. :f) V3 - Since § 4V i-3 7T > 0, these 24. The volume of the trough is maximized when the area of the cross section is maximized. From the diagram the area of the cross section is A(9) = cos 8 + sin 8 cos 8, < 8 < f . Then A' (8) = —sin 8 cos 2 8 — sin 2 ( = - (2 sin 2 6i + sin 8 - 1) sin 8 ^ - 1 when < 8 < there is a maximum. -(2 sin 8 - l)(sin + 1) so A'(6>) = . Also, A'(6>) > for < sin 1 or sin 1 I because < § and A'(0) < for | < < | . Therefore, at Copyright (c) 2006 Pearson Education Section 4.5 Applied Optimization Problems 25 1 25. (a) From the diagram we have: AP = x, RA = y L — x 2 , PB = 8.5 - x, CH = DR=11-RA=11 \/h - x 2 QB = y/x 2 - (8.5 - x) 2 , HQ = 11 -CH-QB = 11 - [ll - VL - x 2 + v/x 2 - (8.5 - x) 2 ] = VL-x 2 - v/x 2 - (8.5 - x) 2 , RQ 2 = RH 2 + HQ 2 = (8.5) 2 + (Vl-x 2 - v/x 2 - (8.5 - x) 2 ) . It follows that RP 2 = PQ 2 + RQ 2 L 2 = x 2 + (Vl 2 - x 2 - v/x 2 - (x - 8.5) 2 ) 2 + (8.5) 2 L 2 - x 2 - ly/l? - x 2 v/17x-(8.5) 2 + 17x - (8.5) 2 + (8.5) 2 L 2 = x 72„2 _ A (1 'I v-'i ; Hv ,',' ^2, I 2 => 17 2 x 2 = 4 (L 2 - x 2 ) (17x - (8.5) 2 ) => L 2 x 2 17¥ (b) Iff(x) 4 [17x-(8.5) 2 ] 17x 3 is minimized, then L 2 is minimized. Now f'(x) Hx- 1 17x- (8.5) 2 4x 3 4x- 17 2X- 1 2X-8.5' 4x-17 4x 2 (8x-51) (4x- 17) 2 f '(x) < when x < f and f ' (x) > when x > 5± . Thus L 2 is minimized when x 51 8 (c) When x '-,thenL« 11.0 in. L 35 30 25 20 15 ^V*' 5 6 — 7 — 6 S lb x 26. (a) From the figure in the text we have P = 2x + 2y => y = | — x. If P = 36, then y = 18 — x. When the cylinder is formed, x = 27rr =>■ r = ^- and h = y => h = 18 — x. The volume of the cylinder is V = 7rr 2 h V(x) 18x 2 -x 3 3x(12-x) 4?r =>• x = or 12; but when x = 0, there is no cylinder. |J =>■ V (12) < =>■ there is a maximum at x = 12. The values of x = 12 cm and . . Solving V'(x) ThenV"(x) = \ (3 ^ ^ v "' y = 6 cm give the largest volume, (b) In this case V(x) = 7rx 2 (18 - x). Solving V'(x) = 37rx(12 - x) = =>• x = or 12; but x = would result in no cylinder. Then V"(x) = 67r(6 — x) =4> V"(12) < => there is a maximum at x = 12. The values of x = 12 cm and y = 6 cm give the largest volume. 27. Note that h 2 3 and so r = a/3 - h 2 . Then the volume is given by V = § r 2 h = |(3 - h 2 )h = Trh - f h 3 for < h < \/3, and so dV it — 7rr 7r(l — r 2 ). The critical point (for h > 0) occurs at h = 1. Since ^ > for < h < 1, and ^ < for 1< h < 3, the critical point corresponds to the maximum volume. The cone of greatest volume has radius y 2 m, height lm, and volume ^ m 3 . 28. (a) f(x) = x 2 (b) f(x) = x 2 f'(x) = x~ 2 (2x 3 - a) , so that f'(x) = when x = 2 implies a = 16 f"(x) = 2x~ 3 (x 3 + a) , so that f"(x) = when x = 1 implies a = -1 29. If f(x) = x 2 + I , then f '(x) = 2x - ax~ 2 and f "(x) = 2 + 2ax~ 3 . The critical points are and 3 ^/| , but x ^ 0. = 6>0 => atx= 3 ,/1j there is a local minimum. However, no local maximum exists for any a. Now f 11 ti 30. If f(x) = x 3 + ax 2 + bx, then f'(x) = 3x 2 + 2ax + b and f"(x) = (a) A local maximum at x = — 1 and local minimum at x = 3 = 27 + 6a + b = =>• a = -3 and b = -9. (b) A local minimum at x = 4 and a point of inflection at x = 1 6x + 2a. , f'(-l) = 0andf'(3) = : => f (4) = and f"(l) = 3 - 2a + b = and 48 + 8a + b = Cfiojt (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle 252 Chapter 4 Applications of Derivatives and 6 + 2a = => a = -3 and b = -24. 31. (a) s(t) = -16t 2 + 96t + 112 =4> v(t) = s'(t) = -32t + 96. At t = 0, the velocity is v(0) = 96 ft/sec. (b) The maximum height ocurs when v(t) = 0, when t = 3. The maximum height is s(3) = 256 ft and it occurs at t = 3 sec. (c) Note that s(t) = -16t 2 + 96t + 112 = -16(t + l)(t - 7), so s = at t = -1 or t = 7. Choosing the positive value of t, the velocity when s = is v(7) = —128 ft/sec. 32. 6 mi ■ Village Jane Let x be the distance from the point on the shoreline nearest Jane's boat to the point where she lands her boat. Then she needs to row y 4 + x 2 mi at 2 mph and walk 6 — x mi at 5 mph. The total amount of time to reach the village is - i. Solving f'(x) = 0, we f(x) have: yT + tp hours (0 < x < 6). Then f'(x) 2 2\/4 + x 2 (2x) 2y/T- 2^/T- 5x = 2^4 + x 2 => 25x 2 = 4(4 + x 2 21x 2 16 ± -$—. We discard the negative \/21 value of x because it is not in the domain. Checking the endpoints and critical point, we have f(0) = 2.2, f( -j— j m 2.12, and f(6) « 3.16. Jane should land her boat -4— w 0.87 miles donw the shoreline from the point nearest her boat. 33. 216 and L(x) 2 h 2 + (x + 27) 2 ,2 + (x+27) 2 is x + 27 "^ " ■ + ^Y + (x + 27)^ when x > 0. Note that L(x) is minimized when f(x) = (8 + -''' minimized. If f'(x) = 0, then 2(8+ 2 ^)(- 2 J#)+2(x + 27)=0 =>■ (x + 27) (l - i|n) = => x = -27 (not acceptable since distance is never negative or x = 12. ThenL(12) = \/2197 « 46.87 ft. 34. (a) From the diagram we have d 2 = 4r 2 — w 2 . The strength of the beam is S = kwd 2 = kw (4r 2 — w 2 ) . When r = 6, then S = 144kw - kw 3 . Also, S'(w) = 144k - 3kw 2 = 3k (48 - w 2 ) so S'(w) = =$> w = ± 4^ ; S" (4^3 ) < and — 4y 3 is not acceptable. Therefore S (4a/ 3 j is the maximum strength. The dimensions of the strongest beam are 4 y 3 by 4 \J 6 inches, (b) (c) 4 6 8 10 12 Both graphs indicate the same maximum value and are consistent with each other. Changing k does not change the dimensions that give the strongest beam (i.e., do not change the values of w and d that produce the strongest beam). Copyright (c) 2006 Pearson Education Section 4.5 Applied Optimization Problems 253 35. (a) From the situation we have w 2 = 144 — d 2 . The stiffness of the beam is S = kwd 3 = kd 3 ( 144 — d 2\l/2 where < d < 12. Also, S'(d) 4kd 2 (108-d 2 ) critical points at 0, 12, and by 3. Both d = and (b) %/144-d 2 d = 12 cause S = 0. The maximum occurs at d = 6 y 3. The dimensions are 6 by 6y 3 inches. (c) "J i 6 S 10 12" Both graphs indicate the same maximum value and are consistent with each other. The changing of k has no effect. => sin t = sin t cos | + sin | cos t =>• sin t = | sin 36. (a) Si = S2 => sin t = sin (t + |) ==> t = f or y" (b) The distance between the particles is s(t) = |si — S2I = sin t— sin (t+ f ) = \ t + Y- cos t ==> tan t = \/3 sin t — v 3 cos t s'(t) t-V3 sin t — v 3 cos t cos t+v^s critical times and endpoints the 2 sin t — \/3 cos t are 0, f , f , f , l -f, 2tt; then s(0) = ^ , s (§) = 0, s (f ) = 1, s (y") = 0, s (±±£) = 1, s(2tt) = ^ greatest distance between the particles is 1. I sin t— v 3 cos t J ( cos t+ y 3 sin 1 1 . (c) Since s (t) = . '—^p 1 we can conclude that at t = f and 3?, s (t) has cusps and 2 sin t - n/3 cos t } J the distance between the particles is changing the fastest near these points. 37. (a) s=10cos(7rt) =>- v = — 107r sin(7rt) =>■ speed = |107r sin (7rt)| = 107r | sin (7rt)| => the maximum speed is 107T w 31.42 cm/sec since the maximum value of |sin (7rt)| is 1; the cart is moving the fastest at t = 0.5 sec, 1.5 sec, 2.5 sec and 3.5 sec when |sin (?rt)| is 1. At these times the distance is s = 10 cos (|) = cm and a = -107r 2 cos(7rt) =>• |a| = 107r 2 | cos (7rt) | => |a| = cm/sec 2 (b) |a| = 10-7T 2 |cos (7rt)| is greatest at t = 0.0 sec, 1.0 sec, 2.0 sec, 3.0 sec and 4.0 sec, and at these times the magnitude of the cart's position is |s| = 10 cm from the rest position and the speed is cm/sec. 38. (a) 2 sin t = sin 2t =>• 2 sin t — 2 sin t cos t = =>• (2 sin t)(l — cos t) = =>• t = kir where k is a positive integer (b) The vertical distance between the masses is s(t) = |sj — s 2 | = ((sj — s 2 ) ) = ((sin 2t — 2 sin t) 2 ) => s'(t) = (i) ((sin 2t - 2 sin t) 2 )~ 1/2 (2)(sin 2t - 2 sin t)(2 cos 2t - 2 cos t) _ 2(cos2t-cost)(sin2t-2sint) _ 4(2 cos t + l)(cos t - l)(sin t)(cos t - 1 ) critical times at |sin 2t— 2 sin t| |sin 2t — 2 sin t| 0, f , 7T, f , 2tt; then s(0) = 0, s (f ) = |sin (f ) - 2 sin (f ) | = ^ , s(tt) = 0, s (f ) sin | -> ) — 2 sin (^) | : s f) 3V3 3V3 T-, s(27r) = => the greatest distance is ^^ at t = % and y 2\l/2 39. (a) s = v/(12- 12t) 2 + (8t) 2 = ((12 - 12t) 2 + 64t (b) | = i ((12 - 12t) 2 + 64t 2 r 1/2 [2(12 - 12t)(-12) + 128t] 208t-144 x /(12-12t) 2 + 64t 2 ds I dt I i=o 1 2 knots and ds I 8 knots Cop # (c) 1 Pearson Ediratio^, k, publishing as Pearson Addison-Wesle 254 Chapter 4 Applications of Derivatives (c) The graph indicates that the ships did not see each other because s(t) > 5 for all values of t. (d) The graph supports the conclusions in parts (b) and (c). (e) lim g t — » 00 at i: m (208t- 144)2 t" 5o 144(1 - 1)2 + 64t2 lim 208 - | t^"So 144 (l-l)%-64 V 144 + 64 which equals the square root of the sums of the squares of the individual speeds 40. The distance OT + TB is minimized when OB is a straight line. Hence Za = Z/3 =£> 9\ = Qi- B A b * -\ \8, si p J JT J • 41. If v = kax — kx 2 , then v' = ka — 2kx and v" = —2k, so v' = v" ( l) = — 2k < 0. The maximum value of v is ^f- . Atx there is a maximum since 42. (a) According to the graph, y'(0) = 0. (b) According to the graph, y'(— L) = 0. (c) y(0) = 0, so d = 0. Now y'(x) = 3ax 2 + 2bx + c, so y'(0) = implies that c = 0. There fore, y(x) = ax 3 + bx 2 and y'(x) = 3ax 2 + 2bx. Then y(-L) = -aL 3 + bL 2 = H and y'(-L) = 3aL 2 - 2bL = 0, so we have two linear equations in two unknowns a and b. The second equation gives b = t=. Substituting into the first equation, we have -aL 3 + as 3aL 3 H, or ^y- = H, so a = 2 ft. Therefore, b = 3ft and the equation for y is y(x)=2fty+3fty,ory(x)=H[2(^) 3 + 3(^) 2 ]. 43. The profit is p = nx - nc = n(x - c) = [a(x - c)" 1 + b(100 - x)] (x - c) = a + b(100 - x)(x - c) = a + (be + 100b)x - lOObc - bx 2 . Then p'(x) = be + 100b - 2bx and p"(x) = -2b. Solving p'(x) = =^> x = | + 50. At x = | + 50 there is a maximum profit since p"(x) = —2b < for all x. 44. Let x represent the number of people over 50. The profit is p(x) = (50 + x)(200 - 2x) - 32(50 + x) - 6000 = -2x 2 + 68x + 2400. Then p'(x) = -4x + 68 and p" = -4. Solving p'(x) = => x = 17. At x = 17 there is a maximum since p"(17) < 0. It would take 67 people to maximize the profit. Copyright (c) 2006 Pearson Education 45. (a) A(q) = kmq : + cm + | q, where q > =^> A'(q) critical points are -J^ , 0, and J ^ , but only W 2 ^ is in the domain. Then A" L/ ^ ] > =• ai Section 4.5 Applied Optimization Problems kmq- 2 + | = ^^ and A"(q) = 2kmq- 3 . The km\ 255 there is a minimum average weekly cost. (b) A(q) (k+bq)m q q = kmq l + bm + cm + | q, where q > => A'(q) = at q 2 ^s as in (a). Also A"(q) = 2kmq 3 > so the most economical quantity to order is still q the average weekly cost. which minimizes 46. We start with c(x) = the cost of producing x items, x > 0, and c(x) to be differentiable. If the average cost can be minimized, it will be at a production level at which dx x c'(x) — c(x) the average cost of producing x items, assumed (by the quotient rule) =>• x c'(x) — c(x) = (multiply both sides by x 2 ) =>• c'(x) = — where c'(x) is the marginal cost. This concludes the proof. (Note: The theorem does not assure a production level that will give a minimum cost, but rather, it indicates where to look to see if there is one. Find the production levels where the average cost equals the marginal cost, then check to see if any of them give a mimimum.) 47. The profit p(x) = r(x) - c(x) = 6x - (x 3 - 6x 2 + 15x) 6x 2 — 9x, where x > 0. Then p'(x) = — 3x 2 + 12x — 9 = — 3(x — 3)(x — 1) and p' (x) = — 6x + 12. The critical points are 1 and 3. Thus p"(l) = 6>0 => atx=l there is a local minimum, and p"(3) = —6 < =>• at x = 3 there is a local maximum. But p(3) = => the best you can do is break even. The average cost of producing x items is c(x) = ^ = x 2 - 20x + 20, 000 => c'(x) = 2x - 20 = =>• x = 10, the only critical value. The average cost is c(10) = $19, 900 per item is a minimum cost because c"(10) = 2 > 0. 49. (a) The artisan should order px units of material in order to have enough until the next delivery. (b) The average number of units in storage until the next delivery is ^ an d so the cost of storing then is s(^) per day, and the total cost for x days is (^)sx. When added to the delivery cost, the total cost for delivery and storage for each cycle is: cost per cycle = d + ^sx. (c) The average cost per day for storage and delivery of materials is: average cost per day = To minimize the average cost per day, set the derivative equal to zero, j- ( d(x)~ + ^x (d+f* 2 ) d X X = - d(x) " + ^=0 => x = ± J |j. Only the positive root makes sense in this context so that x* minimum, check the second derivative ^ ( — d(x)~ + y — . To verify that x* gives a 3 >0 a minimum. The amount to deliver is px* = (d) The line and the hyperbola intersect when 2 = E£x. Solving for x gives Xj ntersect j on = ± » / — . For x > 0, Xintersection = \ if = x*. From this result, the average cost per day is minimized when the average daily cost of delivery is equal to the average daily cost of storage. 50. Average Cost . c« _di dx- 1 (**) x=100 I"' in 100 3 ^ + 96 + 4X 1 / 2 ^> ^ h&\ = -^ + 2X" 1 / 2 = ^> x = 100. Check for a minimum: 100" 3/2 = 0.003 > => a minimum at x = 100. At a production level of 100, 000 units, the average cost will be minimized at $156 per unit. Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 256 Chapter 4 Applications of Derivatives 51. We have || = CM - M 2 . Solving |§ = C - 2M = maximum. M 2 ■ A1SO, dM 3 -2<0=>atM=§ there is ; 52. (a) If v = crgr 2 — cr 3 , then v' = 2cror — 3cr 2 = cr (2tq — 3r) and v" = 2crg — 6cr = 2c (ro — 3r) . The solution of / = is r = or ^ , but is not in the domain. Also, v' > for r < Sfc and v' < for r > ^a at r = ?p there is a maximum. (b) The graph confirms the findings in (a), v oTT O oT3 oTi o.s 53. If x > 0, then (x — l) 2 > =>• x 2 + 1 > 2x =>• ^ti > 2. In particular if a, b, c and d are positive integers, *en (*±i) (Sfi) (*±i) (^i) > 16. 54. (a) f(x) f'(x) ( a 2+x 2 ) 1/2 -x 2 (a 2 +x 2 )- 1/2 .,_ _|_ V Z — V Z v / a 2 + x 2 ~" " w (a 2 + x 2 ) f(x) is an increasing function of x (a 2 + x 2 ) 3/2 (a 2 + x 2 ) 3/2 >0 (b) g(x) ^/b 2 + (d - x) 2 - (b 2 + (d - x) 2 ) + (d - x) 2 ,,, _ - (b 2 + (d - x) 2 ) 1/2 + (d - x) 2 (b 2 + (d - X) 2 )" 1 -' 2 ■ W — b 2 + (d - x) 2 (b 2 + (d - x) 2 ) 3/2 (b 2 + (d - x) 2 ) 3/2 < => g(x) is a decreasing function of x (c) Since ci, C2 > 0, the derivative $- is an increasing function of x (from part (a)) minus a decreasing dx function of x (from part (b)): $■ = — f(x) — — g(x) dx Ci C2 d 2 t _ 1 fl dx 2 c^ f '00 - f g'(x) > since f'(x) > and g'(x) < => j- is an increasing function of x. 55. At x = c, the tangents to the curves are parallel. Justification: The vertical distance between the curves is D(x) = f(x) — g(x), so D'(x) = f'(x) — g'(x). The maximum value of D will occur at a point c where D' = 0. At such a point, f'(c) — g'(c) = 0, or f'(c) = g'(c). 56. (a) f(x) = 3 + 4 cos x + cos 2x is a periodic function with period 2tt (b) No, f(x) = 3 + 4 cos x + cos 2x = 3 + 4 cos x + (2 cos 2 x - 1) = 2 (1 + 2 cos x + cos 2 x) = 2(1 + cos x) 2 > =>• f(x) is never negative 57. (a) If y = cot x — y 2 esc x where < x < it, then y' = (esc x) ( y 2 cot x — esc x j . Solving y' = =>• cos x = -j- => x = I . For < x < | we have y' > 0, and y' < when | < x < n. Therefore, at x = | there is a maximum value of y = — 1 . Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle (b) Section 4.5 Applied Optimization Problems 257 "0T5 I its 1 2T5 5" : The graph confirms the findings in (a). 58. (a) If y = tan x + 3 cot x where < x < | , then y' = sec 2 x — 3 esc 2 x. Solving y' = tan x ±^3 =>• x = ± |, but — I is not in the domain. Also, y" = 2 sec 2 x tan x + 3 esc 2 x cot x > for all < x < | . Therefore at x = | there is a minimum value of y = 2\/3. (b) 80 y = tanx+3cotx 0.25 0.5 0.75 1 1.25 1.5 The graph confirms the findings in (a). 59. (a) The square of the distance is D(x) x- + (s/x + 0) 2 = x 2 - 2x + |, so D'(x) = 2x - 2 and the critical point occurs at x = 1. Since D'(x) < for x < 1 and D'(x) > for x > 1, the critical point corresponds to the minimum distance. The minimum distance is ^/D(l) = %r- . (b) D(x) = x 2 - 2x + ^ 1 4 2.5 2 1.5 >/^^--**> = ^c 1 0.5 1 1 -* 1 1 *■! 0.5 1 1.5 The minimum distance is from the point ( | , 0) to the point ( 1 , 1 ) on the graph of y value x = 1 where D(x), the distance squared, has its minimum value. 60. (a) Calculus Method: The square of the distance from the point ( 1, \J 3 j to ( x, \J 16 — x 2 j is given by D(x) = (x-1) 2 - ^/x, and this occurs at the 16 2x + 1 + 16 - x 2 - 2^48 - 3x 2 + 3 = - 2x + 20 - 2 a/48 - 3x 2 . ThenD'(x) 2- \/48 - 3x 2 (_6x) = - 2 + 0x \/48 - 3x 2 Solving D'(x) = we have: 6x = 2\/48 - 3x 2 Copfiigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 258 Chapter 4 Applications of Derivatives 36x 2 = 4(48 - 3x 2 ) => 9x 2 3x 2 12x 2 ± 2. We discard x -2 as an extraneous solution, leaving x = 2. Since D'(x) < for —4 < x < 2 and D'(x) > for 2 < x < 4, the critical point corresponds to the minimum distance. The minimum distance is ^/D(2) = 2. Geometry Method: The semicircle is centered at the origin and has radius 4. The distance from the origin to I 1 , y3 ) is l 2 + I a/3 ) = 2. The shortest distance from the point to the semicircle is the distance along the radius containing the point I 1, \J 3 ) . That distance is 4 — 2 = 2. (b) The minimum distance is from the point ( 1, y 3 j to the point ( 2, 2y 3 j on the graph of y = \J 16 — x 2 , and this occurs at the value x = 2 where D(x), the distance squared, has its minimum value. 61. (a) The base radius of the cone is r = 2 ™ v ~ an d so me height is h = v a 2 — r 2 = y a- 2n . Therefore, V(x) = fr 2 h=f(^^) ' 27ra - x \ 2 v 2tt ) ■ 3' " 3 ' (b) To simplify the calculations, we shall consider the volume as a function of r: volume = f(r) = ^r 2 y a 2 — r 2 , where < r < a. f ' 2a 2 r - 3r 3 '(r) = l^tVtfTTpi) = | L . i ^= ? (-2r) + (v^^)(2r) -r 3 + 2r(a 2 - r 2 ) \/a 2 — r 2 ^ h= \f a z 7rr(2a 2 - 3r 2 ) 3\/a 2 - r 2 ,2 2a 2 The critical point occurs when r 2 = ^-, which gives r = a-t " a\/6 Then ^-. Using r : n/o and h n/3 we may now find the values of r and h for the given values of a. When a = 4 When a = 5 When a = 6 When a = 8 3 ' 3 ' 5y6 i _ 5y3 , 3 ' " 3 ' r= 21/6^= 2a/3; r _ sy^e h - 5\/3- 1 — 3 , 11 — 3 , (c) Since r = -|- and h = -^-, the relationship is jj = v 2. 62. (a) Let Xo represent the fixed value of x at the point P, so that P has the coordinates (xo, a), and let m = f'(xo) be the slope of the line RT. Then the equation of the line RT is y = m(x — x ) + a. The y-intercept of this line is m(0 — xo) + a = a — mxo, and the x-intercept is the solution of m(x — Xo) + a = 0, or x = m *°~ a . Let O designate the origin. Then (Area of triangle RST) Copffigl (c) 1 Pearson Etation, Inc., publishing as Pearson Addison-Wesle Section 4.5 Applied Optimization Problems 259 2(Area of triangle ORT) 2 • |(x-intercept of line RT)(y-intercept of line RT) (a - mx ) mxp — a ^ / mxp — a ^ 9 1 f mn-o - A 2V m / mx - a \2 V m / Substituting x for Xq, f'(x) for m, and f(x) for a, we have A(x) = — f'(x) M f'(x) (b) The domain is the open interval (0, 10). To graph, let yi = f(x) = 5 + 5i/l — j^, y2 = f'(x) = NDER(yi), and y3 = A(x) = — y2 1 x — — J . The graph of the area function y3 = A(x) is shown below. A(x) 500 400 300 200 100 — < 1 f— 2 4 6 8 10 The vertical asymptotes at x = and x = 10 correspond to horizontal or vertical tangent lines, which do not form triangles. (c) Using our expression for the y-intercept of the tangent line, the height of the triangle is a — mx f(x) -f'(x) -x = 5+ WlOO-x 2 - 5+Wl00-x 2 + - 7 4 i 2^100- 2 2^100 - x 2 2 We may use graphing methods or the analytic method in part (d) to find that the minimum value of A(x) occurs at x w 8.66. Substituting this value into the expression above, the height of the triangle is 15. This is 3 times the y-coordinate of the center of the ellipse, (d) Part (a) remains unchanged. Assuming C > B, the domain is (0, C). To graph, note that 1 f(x)=B + B«/l-£ =B + Pv / C 2 -x 2 andf'(x) B A(x) = -f'(x) M f'(x) Bx cVc 2 - x 2 B + ^v'C 2 -x2 C 2\/C 2 -X 2 2 (-2x) -Bx cVa'- ■ . Therefore we have C\ c 2 -x 2 Bx (BC + B\/C2~ C 2 -x 2 2 C\/C 2 -x 2 Bx BCx\/C 2 - x 2 1 BCxx/C 2 - x 2 Bx 2 + (BC + Ba/C^-x 2 ) (JC 2 - x 2 ) BCx\/C 2 - x 2 Bx 2 + BCa/C 2 - x 2 + B(C 2 - x 2 Bc(c+ v^-x 2 ) BC(C + \/C 2 - x 2 J x\/C 2 - x 2 (xy^^ 2 -)(2)(c + yc 2 ^ 2 -)(-^ ? )-(c +v ^^ 2 -) (x-^+y/c^q)) a v xj - c^ • - x2(c2 - x2) Bc(c + v^-x 2 x 2 (C 2 - x 2 ) -2x 2 - (c + a/c 2 - x 2 x/c 2 " a/C 2 - x 2 ) BC(C + s/&-iA x 2 (C 2 - x 2 ) Cx 2 C^C^ BC(C + \/C 2 -x 2 ) / x 2 (C 2 - x 2 ) (TcJ BC 2 (c+\/C 2 -x 2 ) , ,- ^-^ W i(2x 2 -C 2 -C^ -2x 2 + ^SL^ _ cVc 2 - x 2 + x 2 -(C 2 - x 2 ) (c + \/C 2 - x 2 ) n/c 2 C ') BC X 2( C 2 _ X 2)V 2 Cx 2 - C(C 2 c 2 V^ To find the critical points for < x < C, we solve: 2x 2 - C 2 = cVc 2 - x 2 => 4x 4 - 4C 2 x 2 + C 4 = C 4 - C 2 x 2 =>• 4x 4 — 3C 2 x 2 = =>• x 2 (4x 2 — 3C 2 ) = 0. The minimum value of A(x) for < x < C occurs at the critical point Cop gl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 260 Chapter 4 Applications of Derivatives x = — j- , or x 2 = 2£L. The corresponding triangle height is a — mx = f(x) — f (x) ■ x + y&- -X 2 + Bx 2 Cv / C 2- 3C2 4 + B(3| "J + IVc 2 - X 2 C^- 3C2 4 + §(§) + 4 + B , 3B 2 ' 2 = 3B This shows that the traingle has minimum arrea when its height is 3B. 4.6 INDETERMINATE FORMS AND LHOPITALS RULE 1. l'Hopital: lim x-2 _ 1 7 or lim x-2 lim x-2 lim x " 2 x 2 - 4 2x | x=2 4 U1 x ^" 2 x 2 - 4 ^^ (x - 2) (x + 2) x ^' 2 x + 2 4 2. l'Hopital: lim sin 5x 5 cos 5x lx=0 5 or lim sin 5x 5 lim 5x^0 sin 5x 5x 5-1=5 3. l'Hopital: lim ^ r x — » oo ' lim 10x-3 lim It = I or lim 5x 2 -3x 5x 2 - 3x _ ^^ x — » oo 7x 2 + l x^i'oo 14x x^oo 14 7 """"x^+'oo 7x 2 +l x "^»"bo 7- lim lim 3x- 4. l'Hopital: ^lin^ 4x3 x _ x _ 3 - ^^ T25P^- : n ul x "^j 4x 3 - x -3 - x ^\ ( x -i)(4x 2 +4x+3) or lim lim (x-l)(x 2 + x+l) lim x^ 1 (x 2 + x+l) _ 3 (4x 2 +4x + 3) 11 5. l'Hopital: lim x^O lim x^O 2x lim x^O 2 2 lim "" x , x ^0 x 2 (l + cosx) lim x^O ' sin x \ ( sin x \ ( 1 x / V x / V 1 +cos x/ or lim x^O 1 2 lim x^O (1-cosx) /l + cosx'i x- V 1 + cos X / 6. l'Hopital: lim 1 Y — > r 2x 2 + 3x _ x ^"oo x 3 + x + 1 x — > oo 3x 2 + 1 x lim #H = lim #■ = or lim M+3* oo 6x lim \ -ox 3 + x+l x^ool + -L + 4j 1 7. lim ^ = lim ?^^ = 8. lim 2*=z = lim -?- = 4r = -2 X^tt/2 cosx 0^tt/2 - 8mx - 1 9. lim 5»L| = lim ^ = 4 = 1 10. lim '- sin -* = lim - cosx = lim sinx x^tt/2 1 + cos2x x ^ 7r / 2 -2sm2x x _> ^ 11. lim X — > 7r/4 sin x — cos x lim X — > 7r/4 cos x + sin x V 2 ■ y 2 1 — 2 ' 2 12. lim lim x^tt/3 x ~3 x^tt/3 : -sin x V 3 2 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle i n A Section 4.6 Indeterminate Forms and L'Hopital's Rule 261 13. lim — ( x — 5)tanx= lim x — ^ sin x X — > 7r/2 x^ tt/2 -. (S — x) cos x + sin x(— 1) lim ^ — : x^tt/2 - Slnx 14. lim lim lim 4y^ 4-0 x'^To x + ^v 7 * x _^ 1 + jV x"A"o 2^A+7 " 2-0+7 15. lim x^ 1 2x 2 -(3x + l) % /x + 2 | im 2x^-3^ xV 2 + 2 = Hm 4X -i*f-^ = .J x^ 1 x^ 1 y/^Ts-3 _ ,, m K* 2 + 5 )" 1/2 C 2 *) _ ,;,„ 16. lim V V J ,~ = lim x ^ 2 x2 - 4 x ^ 2 2x lim — = - - x^2 2Vx 2 + 5 6 17. lim x^O ^/a(a + x) -a lim — ^ = -4- x ->0 2Va 2 + ax 2\/a 2 5, where a > 0. 18. Um iO(!HL^i) = Hm lO(cost-l) = lO^sint) = ^ost = ^HM = _ 5 i ' t _» n 3t 2 t-tO 6t t-tO 6 6 3 t-»0 t->0 x(cos X — 1) -xsin x + cqs x — 1 19. lim X ^ C0SX ~ I; = lim x ^q sin x — x x > Q COS X — 1 lim x _, -sin x xcos x — 2sin x \im xcos x + 2sin x ~ x -> sin x lim x -> cos x xsin x + 3 cos x 3 20 H m sin(a + h)-sina _ j^ cos(a + h) -cos a _ g h->0 h->0 21. lim — — =-^ = lim ^j — - = an lim r n 1 = an, where n is a positive integer. r— >l rl r— >!' r — > 1 22. lim x^0+ 23. lim X — > OO fl - 1 ) = hm (i^) = f raop,,al'srule\ = jj m A _ /^ . 1 = V x \/x/ x _> o+ V x / y does not apply y x ^ g+ V V I x - ^^) = x^oo (* - ^^) (xTTlS) _ _ 1 / l'Hopital's rule \ 2 \ is unnecessary y lim x- - ( x^ + x ] lim X — ► OC X + <Jx 2 + X X — > OO x . Vx'^ + x x+^2^ lim — =± x ^°° i + ,/i 24. lim xtan(i) = lim X — » OO V X/ X-KXJ lim X — > OO I lim sec X — » OO 2(t\ sec^O = 1 25. lim 3x-5 lim 26. lim x^0 x^±oo 2x ' x + 2 x ^ ± oo sin7x _ r;™ 7cos(7x) tan llx lim ■1 _ 7 q llsec 2 (llx) 111 11 27. lim ^^ x — > oo v x + 1 lim 9x+l X — > oo x+ 1 lim I = y/9 = 3 x — > oo 1 v 28. lim -)£- x _> 0+ Vsinx lim 512JL VI t^0+ x 1 = 1 29. lim lim x^tt/2" ranx x^tt/2- <• 1 ) I cos X > V cos x / V sin x / lim -4- = 1 x ^ w /2~ Slnx 30. lim lim x ^ + cscx x^0+ (Jrz) x^0+ lim cos x = 1 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 262 Chapter 4 Applications of Derivatives 3 1 . Part (b) is correct because part (a) is neither in the § nor — form and so l'Hopital's rule may not be used. 32. Answers may vary. (a) f(x) = 3x + l;g(x) = x lim f(x) lim 3x + l lim X — > OO gW X — > OG X x^ool ,2 (b) f(x) = x+l;g(x)=x 2 lim -^r = lim x+l lim x'-^J'oo g(x) x^+'oc x 2 x^+'oo 2x ,2. r (c) f(x) = x 2 ;g(x)=x+l lim KM lim -4- lim X — > oo g(x) x->oo x+l x ^ oo 1 OO 33. If f(x) is to be continuous at x = 0, then lim f(x) = f(0) => c = f(0) = lim 9 *- 3s 3 in3 * = lim x — > x — > -' x x — > 9 — 9 cos 3x lim x^O 27 sin 3x 30x lim x^O 31 cos 3x 27 30 ~~ 10 34. (a) For x ^ 0, f'(x) = -f (x + 2) = 1 and g'(x) = -f (x + 1) = 1. Therefore, lim ^1 = 1 = 1, while lim ®& ._ x + 2 _ + 2 _ 9 x+l 0+1 z " (b) This does not contradict l'Hopital's rule because neither f nor g is differentiable at x = (as evidenced by the fact that neither is continuous at x = 0), so l'Hopital's rule does not apply. 35. The graph indicates a limit near — 1 . The limit leads to the indeterminate form § : lim — — p^ — „ . 1 x— l lim x^ l x^ 1 2x 2_ 3x 3/2, x l/2 + 2 _ X- 1 I - 1=1 _ _ ~~ 1 ~~ 4x-|x 1/2 -ix-V2 lim ' — ; — ' X->1 1 2x 2 -(3x + l )Vx+2 36. (a) -0.5 20 40 60 80 100 = X-VX 2 +X (b) The limit leads to the indeterminate form oo — oo lim X — > oo ( x — V x 2 + x J = lim ( x — v x 2 + x ; + \/x 2 +x c + yx 2 +x lim X — » oo / x 2 -(x 2 + x) \ _ V x+\/x 2 + x/ x lim oo x+ y/x 2 + > lim 1 + 1/1 + I l + Vl + x^0 37. Graphing f(x) = - — Sjjjp- on th window [— 1, 1] by [—0.5, 1] it appears that lim f(x) = 0. However, we see that if we let u = x 6 , then lim f(x) = lim i^f^ = lim ^ = 1L,. , x ^ v ' u^0 u u ^ 2u u^O 2 lim 38. (a) We seek c in (-2, 0) so that ( M = ^llV^ = g^f = -\. Since f '(c) = 1 and g'(c) = 2c we have that £ = -\ Cfiojt (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle Section 4.7 Newton's Method 263 (b) We seek c in any open interval (a, b) so that -^ f'(c) _ f(b) - f(a) _ b-a b-a _J_ . J_ _ _J_ (c) g(b)-g(a) b 2 -a 2 (b-a)(b + a) b + a ^ 2c b + a (c) We seek c in (0, 3) so that -4 f'(c) _ f(3)-f(0) -3-0 ■ g(3)-g(0) " 9-0 2c ■\ => 3c 2 + 2c - 12 = => c (Note that c -l-v/37 is not in the given interval (0, 3).) b + a -1+ -y/37 39. (a) By similar triangles, f§ = g§ where E is the point on AB such that CE ± AB : c . B(l,8) \e v^-~^\ /e (x,0) 1 o D 7a(1,0) 1 — x 1 — COS 9 ~ 9 -sin (b) lim (1 - x) Thus ^—q^ = 4 — ^SMr, since the coordinates of C are (cos 0, sin 6). Hence, 1 — x = -4 — ^r^- 6* 9 — sin 9 ' V ' / ' 9 — sin "1-cosfl) |._ 9sin9 + l-cos9 _ i;„, 9 cos g + sin 9 + sin 9 _ i- 9cos9 + 2sin< 8^0 lim lim lim 0^0 n > q 9 — sin 9 rt ^q 1 — cos 9 sin 9) + cos 9 + 2cos 9 _ ,. -9sin9 + 3cos9 _ + 3 cos B n . Q cos 1 lim 8^0 sin 9 lim 8^0 sin 9 (c) We have that lim [(1 - x) - (1 - cos 0)1 = lim 9 — > oo 9 — » oc - (1 -COS0) lim (1 — cos 9) 9 — » oo As — ► oo, (1 — cos 8) oscillates between and 2, and so it is bounded. Since lim 8 — > oo 9 - sin 9 — J ) - 1—1 = - 1 lim (1 — cos 6) 8 — > oo approaches 0. - 1 0. Geometrically, this means that as 6 — » oo, the distance between points P and D 40. Throughout this problem note that r 2 = y 2 + 1, r > y and that both r — > oo and y — » oo as (a) lim r — y = (b) n lim r 2 — y 2 lim -i- = -tt/2 r +y lim 1 = 1 9 -» tt/2 6»^tt/2 (c) We have that r 3 - y 3 = (r - y)(r 2 + ry + y 2 ) r + ry + y - y + y-y + y r + y ■* r 3y : f- = 3y?. Since lim 3y 9 -> tt/2 lim 3sin # • y = oo we have that lim r 3 — y -» tt/2 6»^tt/2 4.7 NEWTON'S METHOD i- y x 2 *2 1 =>. y' = 2x + 1 t+I-l 7 > x 2 = i 2x„ + l ' X 9 _ 4-2-1 ^ A i 1 2 _ 4+6-9 _ ? 12+9 -1.66667 21 13 21 1 => Xi 61905; x = l+i-i 2+1 ' X : = 1-1-1 -2+1 3x + 1 =4> y' = 3x 2 + 3 =4> x n+ x 2 -l+i J+3 1 + 1 90 29 90 xg+3x„+l . 3x2+3 ' X -0.32222 => Xj = Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 264 Chapter 4 Applications of Derivatives 3. y = x 4 + x - 3 => y' = 4x 3 + 1 =4> x n+1 = x n 1296 , 6 o 625 ~t~5 J ^±^^;x =l => x 1 = l i±l=2 Y _ 6 X 2 - 5 1296+750-1875 -2 => x 2 = -2 - + 1 5 4320+625 16-2-3 __ 2+ l -32+1 4x3+1 _ 6 _ J71_ _ 5763 r 5 4945 4945 " -§} rs -1.64516 4+1 1.16542; x = -1 => Xj = -1 - 1-1-3 2x n -x2+l 4. y = 2x - x 2 + 1 => y' = 2 - 2x => x n+1 = x„ - ^-2x7' ; x o = ° ^ x i = ° ~'~' +1 -_l+i=_i« -.41667; x = 2 => xi = 2 4 ~ 4+1 0-0+1 2-0 X 2 2 2+1 5 _ 20-25+4 _ 5 _ 1 -12 2 12 2 ' 12 12 § « 2.41667 2-4 1 2 X 2 5-f+l 2-5 5. y = x 4 - 2 =>• y' = 4x 3 =^ x 5 _ 113 _ 2500-113 _ 2387 ~~ 4 2000 ~~ 2000 ~~ 2000 \ 4 - 4x2 " -;x = l =^ x x = 1- V *2 5 _ 625-512 4 2000 1.1935 6. From Exercise 5, x n x 4 — 2 . ,1-2 x » - ip- ; x o = -i => xi = -l - y X 2 = - 7 5 _ 625-512 4 -2000 113 2000 1.1935 7. f(x ) = and f (x ) ^ f(x„ v„+l -a." f77x"J gives X! = x => x 2 = x the approximations in Newton's method will be the root of f(x) = 0. x n = Xq for all n > 0. That is, all of 8. It does matter. If you start too far away from x = | , the calculated values may approach some other root. Starting with Xq = —0.5, for instance, leads to x = — | as the root, not x = | . 9. Ifx = h>0 /h Xi =x f(xo) f'(x ) h f(h) f'(h) = h ifx = -h<0 h - (v/h) (2>/h) = -h; ^ x i= x o-^ = -h- = -h+ (Vh) h-yfh) =h. f(-h) f'(-h) ~{i ,x>0 ■4^x,x<0 10. f(x) = x 1 / 3 => f'(x) -2/3 ,1/3 Q): -2/3 = — 2x n ; Xo = 1 =>• Xi = —2, x 2 = 4, X3 = —8, and X4 = 16 and so forth. Since |x n | = 2|x„_,| we may conclude that n — > 00 =>• IxJ — > 00. 11. i) is equivalent to solving x 3 — 3x — 1 = 0. ii) is equivalent to solving x 3 — 3x — 1 = 0. iii) is equivalent to solving x 3 — 3x — 1 = 0. iv) is equivalent to solving x 3 — 3x — 1 = 0. All four equations are equivalent. 12. f(x) = x - 1 - 0.5 sin x =^> f'(x) =1-0.5 cos x X! = 1.49870 1 — 1 — 0.5 sin x. 1 — 0.5 cos x„ - ;if x = 1.5, then Cop # (c) 1 Pearson link, Ire, publishing as Pearson Addison-Wesle 13. For Xq = —0.3, the procedure converges to the root —0.32218535. (a) Plotl Plots PlotJ \ylBx A 3+3x+l ■*.y2BnDer(yl ix.x) r. I V I INSf I DELf ISELCTt (b) '.3-*x (C) x-al/y2-»x -.322324152194 -.322185360292 -.322135354626 -.322135354626 (d) Values for x will vary. One possible choice is Xq = 0.1. x-yl^y2-*x .1 . 329372795537 -.322200595043 -.322135354698 -.322135354626 (e) Values for x will vary. Section 4.7 Newton's Method 265 14. (a) f(x) = x 3 - 3x - 1 =>• f'(x) = 3x 2 - 3 => x n+I and -0.34730 (b) The estimated solutions of x 3 — 3x — 1 = are -1.53209,-0.34730, 1.87939. x; - 3x„ - l 3*?-3 the two negative zeros are —1.53209 (c) The estimated x-values where g(x) = 0.25x 4 — 1 .5x 2 — x + 5 has horizontal tangents are the roots of g'(x) = x 3 — 3x — 1, and these are -1.53209,-0.34730, 1.87939. -2/ -1 ffx) = x 3 -3x-l g(x) = 0.25x 4 -1.5x 2 -x + 5 15. f(x) = tan x - 2x => f'(x) = sec 2 x - 2 => x n+1 = x„ => x 2 = 1.155327774 => x 16 = x 17 = 1.165561185 tan(x„)-2x„ sec 2 (x n ) ; x = 1 => xi = 12920445 16. f(x) = x 4 - 2x 3 - x 2 - 2x + 2 =4> f'(x) = 4x 3 - 6x 2 - 2x - 2 => x„ ifx = 0.5, then x 4 = 0.630115396; if x = 2.5, then x 4 = 2.57327196 X n ZX n X n ZX n + L n n n Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 266 Chapter 4 Applications of Derivatives 17. (a) The graph of f(x) = sin 3x — 0.99 + x 2 in the window — 2 < x < 2, —2 < y < 3 suggests three roots. However, when you zoom in on the x-axis near x = 1.2, you can see that the graph lies above the axis there. There are only two roots, one near x = — 1 , the other near x = 0.4. (b) f(x) = sin 3x - 0.99 + x 2 =>■ f ' (x) = 3 cos 3x + 2x and the solutions sin(3x„) - 0.99+x 2 -v a, 1+1 - ^ n 3 cos(3Xn ) + 2x„ are approximately 0.35003501505249 and -1.0261731615301 y - sln(3x) - 0.99 + x 18. (a) Yes, three times as indicted by the graphs (b) f(x) = cos 3x - x =>■ f ' (x) = —3 sin 3x — 1 => x n+ . cos(3x n )-x n ; at 11 —3 sin (3x n ) — 1 approximately —0.979367, -0.887726, and 0.39004 we have cos 3x = x 19. f(x) = 2x 4 - 4x 2 + 1 =>• f'(x) = 8x 3 _ 2x;j-4x 2 +l .- X » 8x3 _ 8x . II X -2, thenx G = -1.30656296; if Xo = —0.5, then X3 = —0.5411961; the roots are approximately ± 0.5411961 and ± 1.30656296 because f(x) is an even function. 20. f(x) = tan x => f '(x) = sec 2 x approximate -it to be 3.14159. tan (x n ) sec 2 (x„) ;x xi = 3.13971 =* x 2 = 3.14159 and we 21. From the graph we let Xo = 0.5 and f(x) = cos x — 2x =* x "+' = x » " -li^J-^ => x i = - 45063 =^> X2 = .45018 =^> at x « 0.45 we have cos x = 2x. y 3 2 / y = 2x -3 yf -1 J 1 \2 3 — ^ /l y = cos x / / " 2 ' -3 22. From the graph we let x = —0.7 and f(x) = cos x + x =► X »+> = X » " T^SxT =► * = - 73944 =^ X2 = —.73908 => at x ss —0.74 we have cos x = — x. 3 ' \y = -x 2 ~">. y = cos x -3 -/ -1 -1 -2 \1 N2 3 -3 Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 4.7 Newton's Method 267 by the Intermediate Value Theorem the equation x^ + 2x n -4 23. If f(x) = x 3 + 2x - 4, then f(l) = - 1 < and f(2) = 8 > x 3 + 2x — 4 = has a solution between 1 and 2. Consequently, f'(x) = 3x 2 + 2 and x n+1 -a,- - - , , ,— . Thenx =1 =>• x x = 1.2 => x 2 = 1.17975 =>• x 3 = 1.179509 =>• x 4 = 1.1795090 =>• the root is approximately 1.17951. 24. We wish to solve 8x 4 - 14x 3 - 9x 2 + 1 lx - 1 = 0. Let f(x) = 8x 4 - 14x 3 - 9x 2 + llx - 1, then .3 ^-.2 io_ , ,, 8x4 -14x3- 9x2 + llx„-l f (x) = 32x 3 - 42x 2 - 18x + 1 1 32x- ! -42x 2 -18x n + ll x approximation of corresponding root -1.0 -0.976823589 0.1 0.100363332 0.6 0.642746671 2.0 1.983713587 25. f(x) = 4x 4 - 4x 2 f'(x) = 16x 3 - 8x f(x.) POO Ix? - . Iterations are performed using the procedure in problem 13 in this section. (a) For Xo = 2 or Xo = —0.8, x; — > — 1 as i gets large. (b) For Xo = —0.5 or Xo = 0.25, x; — > as i gets large. (c) For Xo = 0.8 or Xo = 2, x; — > 1 as i gets large. (d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal value.) Forx = -^orx = -^, /21 Xo ^orx Newton's method does not converge. The values of x; alternate between as l increases. 26. (a) The distance can be represented by D(x) = J(x - 2) 2 + (x 2 + i) 2 , where x > 0. The distance D(x) is minimized when f(x) = (x - 2) 2 + (x 2 + |) 2 is minimized. If f(x) = (x - 2) 2 f'(x) = 4 (x 3 + x - 1) and f"(x) = 4 (3x 2 + 1) > 0. Now f'(x) = (x 2 + i) 2 ,then x- 1 =0 =>■ x(x 2 + 1) = 1 X 2+l (b) Letg(x) X 2+l (x 2 + 1)- ?'(x) (x 2 + 1) z (2x) - 1 -2x W ;x = l (x 2 +l) 2 X4 = 0.68233 to five decimal places. 27. f(x) = (x - 1) 4U => f'(x) = 40(x - l) 3y => x n x„ — 40(x n -l) 3 39x„ + 1 40 With x = 2, our computer jave x 87 = x 88 = x 89 x 200 = 1 ■ 1 105 1 , coming within 0. 1 105 1 of the root x = 1 . "n (K ~ 1) 28. f(x) = 4x 4 - 4x 2 => f'(x) = 16x 3 - 8x = 8x (2x 2 - 1) =4> x n+1 = x„ - ^ _ ^ ; if x = .65, then X12 « -.000004, if x = .7, then x 12 = -1.000004; if x = .8, then x 6 = 1.000000. NOTE: ^ « .654654 29. f(x) = x 3 + 3.6x 2 - 36.4 =4> f'(x) = 3x 2 + 7.2x =>■ x„ xj + 3.6x 2 - 36.4 ; x = 2 => xi = 2.5303 »+'- A " 3x2 + 7.2x n x 2 = 2.45418225 =>• x 3 = 2.45238021 =>• x 4 = 2.45237921 which is 2.45 to two decimal places. Recall that Cop # (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 268 Chapter 4 Applications of Derivatives x = 10 4 [H 3 0+] => [H 3 0+] = (x) (10~ 4 ) = (2.45) (1(T 4 ) = 0.000245 30. Newton's method yields the following: the initial value 2 i V^ + i the approached value 1 -5.5593H -29.5815 - 17.0789i 4.8 ANTIDERIVATIVES 1. (a) X 2 2. (a) 3x 2 3. (a) x- 3 4. (a) -x- 2 5. (a) X 6. (a) 1 X 2 7. (a) v 7 ^ 8. (a) x 4/3 9. (a) x 2/3 10. (a) x l/2 11. (a) COS (7TX) 12. (a) sin (7rx) 13. (a) tan x 14. (a) —cot X 15. (a) —esc X 16. (a) sec x 17. J(x+ l)dx= f +x + C 19. J(3t 2 + |)dt = t 3 + f +C (V) 3 (b) X s 8 (b) X" 3 3 (b) X -2 1 X" 4 ~r 3 (b) ^5 X (b) -1 4x2 (b) v^ (b) 1 Y 2/3 2 X (b) x l/3 (b) x -l/2 (b) —3 cos x (b) sin(f) (b) 2 tan (f) (b) cot(f) (b) j esc (5x) (b) | sec(3x) (C) y - X 2 + X (c) f - 3x 2 + 8x (c) - ^ + x 2 + 3x (C) V + T 5 (c) 2x (c) 4 ^ 2x2 (c) fv^ + 2^ ( C ) 3 x 4/3 + 3 x 2/3 (c) x" 1 / 3 (c) x- 3 / 2 (c) —COS (7TX) cos(3x) (c) (f)sin(f)+ 7 r l (c) - § tan (|) (c) x + 4cot(2x) (c) 2csc(f) (c) fsec(f) 18. / (5 - 6x) dx = 5x - 3x 2 + C 20. J U + 4t 3 )dt=£ + t 4 + C Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 4.8 Antiderivatives 269 21. J (2x 3 - 5x + 7) dx = i x 4 - § x 2 + 7x + C 22. J" (1 - x 2 - 3x 5 ) dx = x - | x 3 - i x G + C 23- J(^-x 2 -i)dx=/(x- 2 -x 2 -i; dx -13 3 x + C 2x 2 1_ X^ X x 3 3 ^ +C ==- + ^+X Z + C c = -^ + c 24- / Q - Jr +2x) dx = / (i - 2x- 3 + 2x) dx = ± x - (^) 25. fx- 1 /3 d x=^+C=|x 2 / 3 +C 26. f x - 5 / 4 dx=^ + 27. / (^ + Vx) dx = / (x 1 / 2 + x 1 / 3 ) dx =^ + ^ + C=? x 3 / 2 + | x 4 / 3 + C 28. / (^ + -^) dx = / (i x 1 / 2 + 2X- 1 / 2 ) dx = 1 (^) + 2 (*£) + C = | x 3 / 2 + 4X 1 / 2 + C 29. /(8y-^)dy = /(8y-2y- 1 / 4 )dy=¥-2(f)+C = 4y 2 -fy 3 / 4 + C 30. /(*-£) dy = /($-r^ 4 )dy = *y-(^)+C = $ + £ + C 31. / 2x (1 - x- 3 ) dx = / (2x - 2x~ 2 ) dx = ^ - 2 fe) C = x 2 i 2 32 x 2x 2 + *~ J x- 3 (x + 1) dx = J (x- 2 + x- 3 ) dx = £i + f il) + C 33. /^dt = /(¥ + ¥)*=/(f 1/s +r' /!, )*=^+(^)+c = 2 > /t 34. J^dt=/(i + ^)dt=/(4t- 3 + t-/ 2 )dt = 4(^) + (^) + C * +C t 2 3t 3/2 T «- 35. J" -2 cos t dt = -2 sin t + C 37. /7sin | d0 = -21 cos f + C 39. J" -3 esc 2 x dx = 3 cot x + C 41. j cscg 2 cotfl d0 = - \ csc 6 + C 43. J (4 sec x tan x — 2 sec 2 x) dx = 4 sec x — 2 tan x + C 36. J -5 sin t dt = 5 cos t + C 38. J 3 cos 50 d0 = | sin 50 + C 40. / - ^ dx = - ^p + C 42. / § sec 6> tan d0 = | sec (9 + C 44. J i (csc 2 x — csc x cot x) dx = — i cot x + | csc x + C 45. J (sin 2x — csc 2 x) dx = — \ cos 2x + cot x + C 46. J (2 cos 2x — 3 sin 3x) dx = sin 2x + cos 3x + C 47. / l±s&* dt = / (I + | cos 4t) dt = \ t + \ (^) + C = \ + ^ + C 48. / Mf^ dt = / (i - i cos fit) dt = 1 1 - 1 (^S) + C = \ - ^ + C Copffigl (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 270 Chapter 4 Applications of Derivatives 49. / (1 + tan 2 0) 6.9 = / sec 2 Ad = tan + C 50. / (2 + tan 2 9) d0 = f (1 + 1 + tan 2 0) d0 = / (1 + sec 2 0) d0 = + tan + C 51. J cot 2 x dx = J (esc 2 x — 1) dx = — cot x — x + C 52. f (1 - cot 2 x) dx = J (1 - (esc 2 x - 1)) dx = J" (2 - esc 2 x) dx = 2x + cot x + C 53. / cos (tan + sec 0) d0 = / (sin + 1) d0 = -cos + + C 54. f f cS . , d0 = f ( f cB . g ) (*Pf) d0 = f t-4-j, d0 = f ^j-j d0 = f sec 2 d0 = tan i J csc 6 — sin J V esc 6 — sin / V sin ) J 1 —sin 2 6 J cos 2 9 J 55- ^( (Z ^+c) = 4 ^#^=(7x-2)3 56- £ (_ i^x+^i + ^ = _ £_ (3» + 5)-'(3) j = (3x + 5) . 57. £ (i tan(5x - 1) + C) = ± (sec 2 (5x - 1)) (5) = sec 2 (5x - 1) 58. 59. dx (_3 cot (i=i) + C) = -3 (-csc 2 (s=i)) (i) = csc 2 (s=i) dx V x i + c) = (-i)(-i)( X +ir (X+1)2 uu - dx Vx+1 c) (x+l)(l)-x(l) (X + 1)2 (X+l)2 sin x + y cos x = x sin x + y cos x 7^ x sin x 61. (a) Wrong: £ (f sin x + c) = (b) Wrong: £ (— x cos x + C) = —cos x + x sin x ^ x sin x (c) Right: gj (— x cos x + sin x + C) = —cos x + x sin x + cos x = x sin x 62. (a) Wrong: £ (^ + c) = 3 - J f^ (sec tan 0) = sec 3 tan ^ tan sec 2 (b) Right: £ (i tan 2 + C) = 1 (2 tan 0) sec 2 = tan sec 2 (c) Right: ^ (1 sec 2 + C) = \ (2 sec 0) sec tan = tan sec 2 63. (a) Wrong: £ (2* + i£ + c) = 3(2x + 1)2(2) = 2(2x + l) 2 / (2x + l) 2 (b) Wrong: £ ((2x + l) 3 + C) = 3(2x + 1) 2 (2) = 6(2x + l) 2 f 3(2x + l) 2 (c) Right: £ ((2x + l) 3 + C) = 6(2x + l) 2 64. (a) Wrong: £ (x 2 + x + cf' 2 = \ (x 2 + x + C) 1/L (2x + 1) = 2y/ ^ + c + ^2x + 1 (b) Wrong: £ ((x 2 + x) 1/2 + c) = * (x 2 + x)~ 1/2 (2x + 1) = ^= ^ ^/2x + 1 (c) Right: U (v/2x+l) 3 + CJ = £ Q (2x + I) 3 / 2 + C) = § (2x + l) 1 /2(2) = V / 2xTT 65. Graph (b), because ^ = 2x => y = x 2 + C. Then y(l) = 4 => C = 3. 66. Graph (b), because % = -x =>• y = - | x 2 + C. Then y(-l) = 1 => C = | Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 4.8 Antiderivatives 271 67. dy 2x - 7 =>■ y = x 2 - 7x + C; at x = 2 and y = we have = 2 2 - 7(2) + C => C = 10 => y = x 2 - 7x + 10 68. dy 10 -x => y = lOx- \ +C;atx = Oandy = -1 we have -1 = 10(0)- \ +C =>• C = 1 =$► y = 10x - \ - 1 69. & - 1 dx x 2 x 1 + y+C;atx = 2 and y = 1 we have 1 ■U? + c y = — x * + y — i or y 1 , xf _ 1 x + 2 2 70. ^=9x 2 -4x + 5 => y = 3x 3 - 2x 2 + 5x + C; at x = -1 and y = we have = 3(-l) 3 - 2(-l) 2 + 5(-l) + C => C = 10 =>• y = 3x 3 - 2x 2 + 5x + 10 71. % = 3x- 2 / 3 => y=3^+C = 9;atx = 9X 1 / 3 + C; at x = -1 and y = -5 we have -5 = O(-l) 1 / 3 + C => C = 4 d\ :ix' ■' l 3 y = 9X 1 / 3 + 4 72. %- = -X- = \ x- 1 / 2 => y = x 1 / 2 + C; at x = 4 and y = we have = 4 1 / 2 + C^C = -2^y = x 1 / 2 - 2 73. jjj = 1 + cos t => s = t + sin t + C; at t = and s = 4 we have 4=0+sinO + C => C = 4 =$■ s = t+sint+4 74. gj = cos t + sin t => s = sin t — cos t + C; at t = it and s = 1 we have 1 = sin n — cos tt + C => C = =$> s = sin t — cos t 75. g = -7r sin 7t6» =>■ r = cos(7T0) + C; at r = and = we have = cos(7r0) + C => C = -1 => r = cos(7T0) - 1 76. % = cos tt0 => r = \ sin(7T0) + C; at r = 1 and = we have 1 = \ sin (vrO) + C=^C=l=^r=isin (tt0) + 1 77. dv _ 1 t tan t => v = \ sec t + C; at v = 1 and t = we have 1 = \ sec (0) + C => C = \ =4> v = \ sec t + \ 78. dv ,2 J( - 8t + esc 2 1 => v = 4t 2 - cot t + C; at v = -7 and t = f we have -7 = 4 (| ) - cot (§) + C =>• C = — 7 - 7r =>• v = 4t 2 - cot t - 7 - 7T 2 d^ Jy dy 79. H = 2 - 6x ^> g = 2x - 3x 2 + Ci; at §- = 4 and x = we have 4 = 2(0) - 3(0) 2 + Ci => Ci = 4 =4> g = 2x - 3x 2 + 4 =>• y = x 2 - x 3 + 4x + C 2 ; at y = 1 and x = we have 1 = 2 - 3 + 4(0) + C 2 => C 2 = 1 =>■ y = x 2 - x 3 + 4x + 1 80. g = => | = Ci; at ^ = 2 and x = we have d = 2 => g = 2 =4> y = 2x + C 2 ; at y = and x = we have = 2(0) + C 2 => C 2 = => y = 2x 81. gf = 2 =2 r 3 ^ | = -r 2 + Ci;at| = landt= 1 we have 1 = -(1)~ 2 + Ci => Ci = 2 => g = -r 2 + 2 => r = r 1 + 2t + C 2 ; at r = 1 and t = 1 we have 1 = l -1 + 2(1) + C 2 => C 2 = -2 => r = t" 1 + 2t - 2 or r = ± + 2t - 2 82. d 2 s _ 3t . ds _ 3t 2 W-J => | = t+ C i; at l = 3andt = 4wehave3 =T? L + C i =* C i=° =* l = f => s=^+C 2 ;at s = 4 and t = 4 we have 4 = ^ + C 2 => C 2 = =^ s = ^ Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 272 Chapter 4 Applications of Derivatives 83- g=6 g = 6x + Ci; at = -8 and x = we have -8 = 6(0) + d => d = -8 — — g = 3x 2 - 8x + C 2 ; at g = and x = we have = 3(0) 2 - 8(0) + C 2 3 _ Asr'2 i r„- ot u - s onH v — n ,1,0 havo ^ — n3 _ 4 cm 2 1 r". -^ r^ — ^ -■*. „ _ v 3 /i-»2 dx2 dy 6x-8 C 2 = =>- g = 3x 2 - 8x =4> y = X J - 4x 2 + C 3 ; at y = 5 and x = we have 5 = 3 - 4(0) 2 + C 3 => C 3 = 5 => y = x 3 - 4x 2 + 5 84. ff = => ff = d; at ff = -2 and t = we have |f = -2 => f = -2t + C 2 ; at f = - \ and t = we have - \ = -2(0) + C 2 => C 2 = - | - - dfl - ^ 1 — fl - * 2 4 d» dt -2t-i => r - 1 1 + C 3 ; at 9 = yjl and t = we have 'I = -0 2 - | (0) + C 3 => C 3 = \/2 => 6» = -t 2 - i t + V2 85. y (4) = -sin t + cos t =>- y'" = cos t + sin t + Ci; at y'" = 7 and t = we have 7 = cos(0) + sin(0) + Ci => Ci = 6 => y'" = cos t + sin t + 6 => y" = sin t - cos t + 6t + C 2 ; at y" = - 1 and t = we have -1 = sin(0) - cos (0) + 6(0) + d =^ C 2 = =>• y" = sin t - cos t + 6t => y' = -cost- sint+ 3t 2 + C 3 ; at y' = - 1 and t = we have - 1 = -cos (0) - sin (0) + 3(0) 2 + C 3 => C 3 = =4> y' = -cos t - sin t + 3t 2 =4> y = -sin t + cos t + t 3 + C 4 ; at y = and t = we have = -sin (0) + cos (0) + 3 + C 4 => C 4 = - 1 =4> y = —sin t + cos t + t 3 — 1 86. y' 4 ' = —cos x + 8 sin (2x) => y'" = — sin x — 4 cos (2x) + Ci; at y'" = and x = we have = -sin (0) - 4 cos (2(0)) + Ci =4> Ci = 4 =>■ y'" = -sin x - 4 cos (2x) + 4 => y" = cos x - 2 sin (2x) + 4x + C 2 ; at y" = 1 and x = we have 1 = cos (0) - 2 sin (2(0)) + 4(0) + C 2 => C 2 = => y" = cos x - 2 sin (2x) + 4x => y' = sin x + cos (2x) + 2x 2 + C 3 ; at y' = 1 and x = we have 1 = sin (0) + cos (2(0)) + 2(0) 2 + C 3 => C 3 = => y' = sin x + cos (2x) + 2x 2 => y — —cos x + i sin (2x) + § x 3 + C 4 ; at y = 3 and x = we have 3 = -cos (0) + \ sin (2(0)) + § (0) 3 + C 4 =^C 4 = 4=^y = -cos x + \ sin (2x) + § x 3 + 4 87. m = y' = 3Jx = 3x 1/2 => y = 2x 3 / 2 + C; at (9,4) we have 4 = 2(9) 3/2 + C => C = -50 =4> y = 2x 3 / 2 - 50 I (a) g = 6x ^ = 3x 2 + Ci; at y' = and x = we have = 3(0) 2 + d => d = l=3x 2 => y = x 3 + C 2 ; at y = 1 and x = we have C 2 = 1 => y = x 3 + 1 (b) One, because any other possible function would differ from x 3 + 1 by a constant that must be zero because of the initial conditions 89. g = 1- f x 1 / 3 => y = J(l - f-x 1 / 3 ) dx = x-x 4 / 3 + C; at (1,0.5) on the curve we have 0.5 = 1 - l 4 / 3 + C ^ C = 0.5 => y = x- x 4 / 3 + \ 90. g =x- 1 =>■ y = J(x- l)dx= ^ -x + C;at(-l,l)onthecurvewehave 1 = ^ -(-1) + C =* c = -i =► y=f- x -5 91. g| = sin x — cos x =$■ y = J (sin x — cos x) dx = —cos x — sin x + C; at (— tt, —1) on the curve we have — 1 = — cos(— 7r) — sin(— 7r) + C => C = — 2 =4- y = — cos x — sin x — 2 92. ^ = ^k- + tt sin 7rx = \ x~ 1/2 + 7r sin 7rx => y = / (| x~ 1/2 + sin 7rx) dx = x 1/2 - cos 7rx + C; at (1,2) on the curve we have 2 = l 1 / 2 — cos 7r(l) + C => C = =4- y= \fx — cos 7rx 93. (a) f = 9.8t - 3 =4> s = 4.9t 2 - 3t + C; (i) at s = 5 and t = we have C = 5 => s = 4.9t 2 - 3t + 5; displacement = s(3) - s(l) = ((4.9)(9) - 9 + 5) - (4.9 - 3 + 5) = 33.2 units; (ii) at s = -2 and t = we have C = -2 =>■ s = 4.9t 2 - 3t - 2; displacement = s(3) - s(l) = ((4.9)(9) - 9 - 2) - (4.9 - 3 - 2) = 33.2 units; (iii) at s = So and t = we have C = so =>- s = 4.9t 2 — 3t + So; displacement = s(3) — s(l) = ((4.9)(9) - 9 + s ) - (4.9 - 3 + s ) = 33.2 units Copffigl (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 4.8 Antiderivatives 273 (b) True. Given an antiderivative f(t) of the velocity function, we know that the body's position function is s = f(t) + C for some constant C. Therefore, the displacement from t = a to t = b is (f(b) + C) — (f(a) + C) = f(b) — f(a). Thus we can find the displacement from any antiderivative f as the numerical difference f(b) — f(a) without knowing the exact values of C and s. 94. a(t) = v'(t) = 20 = 1200 m/sec. v(t) = 20t + C; at (0, 0) we have C = => v(t) = 20t. When t = 60, then v(60) = 20(60) 95. Step 1: p = -k => f = -kt + d; at f = 88 and t = we have d s = -k (| J + 88t + C 2 ; at s = and t = we have C 2 = => ds dt -kt + : Step 2: ^=0 Step 3: 242 = = -kt + 88 2 > t = 242 88 k (W , (88)1 2k ' k 242 2k + 88t 16 96. eft dt- ds dt / -k dt = -kt + C; at | = 44 when t = we have 44 = -k(0) + C =>■ C = 44 ds dt kt 2 kt + 44 =$> s = - s- + 44t + Ci; at s = when t = we have kfO)' 44(0) + Ci Ci =0 f + 44t. Then f =>■ -kt + 44 = => t fands(f) ^r + 44(f 45 968 , 1936 45 «6s k 45 968 21.5 ft sec 2 97. (a) v = / a dt = j (l5t J / 2 - 3t- J / 2 ) dt = 10t 3 / 2 - 6I 1 / 2 + C; f (1) = 4 =^ 4= 10(1) 3 / 2 - 6a) 1 / 2 + C => C = =► v = 10t 3 / 2 - 6tV 2 (b) s = / v dt = J*(l0t 3 / 2 - 6t J / 2 ) dt = 4t 5 / 2 - 4t 3 / 2 + C; s(l) = => = 4(1) 5 / 2 - 4(1) 3 / 2 + C => C = =>. s = 4t 5 / 2 - 4t 3 / 2 eft dt 2 -5.2 i = -5.2t + Ci; at | = and t = we have Ci = ds dt -5.2t => s = -2.6t 2 + d; at s = 4 and t = we have Co -2.6t 2 + 4. Then s = => = -2.6t 2 + 4 => t 1.24 sec, since t > 99. eft dt- => | = J a dt = at + C; | = v when t = =^> C vn |=at + v v t + Ci;s = s when t = „ _ a(0) 2 s — -^T v (0) + Ci =>• d = so f + v t + s 100. The appropriate initial value problem is: Differential Equation: dt 2 -g with Initial Conditions: Vq and s = Sq when t = 0. Thus, J-gdt ? t + C i; |(0) = v => v = (-g)(0) + Ci =► Ci vo =► i = -8t + v . Thus s = /(-gt + v ) dt = - 1 gt 2 + v t + C 2 ; s(0) = s =-■ - | (g)(0) 2 + v (0) + C 2 =>• C 2 = s Thus s i gt 2 + v t + s . 101. (a) Jf(x)dx= 1- v/x + d = -v/x + C (b) Jg(x) dx = x + 2 + Ci = x + C (c) J-f(x) dx = - (1 - y/x) + d = y/x+ C (d) J-g(x) dx = -(x + 2) + Ci = -x + C (e) / [f(x) + g(x)] dx = (1 - yfi) + (x + 2) + Ci = x - y/i + C (f) / [f(x) - g(x)] dx = (1 - y/x) - (x + 2) + Cx = -x - yfi. + C 102. Yes. If F(x) and G(x) both solve the initial value problem on an interval I then they both have the same first derivative. Therefore, by Corollary 2 of the Mean Value Theorem there is a constant C such that F(x) = G(x) + C for all x. In particular, F(x ) = G(x ) + C, so C = F(x ) - G(x ) = 0. Hence F(x) = G(x) for all x. Cop # (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 274 Chapter 4 Applications of Derivatives 103 — 106 Example CAS commands: Maple : with(student): f := x -> cos(x) A 2 + sin(x); ic := [x=Pi,y=l]; F := unapply( int( f(x), x ) + C, x ); eq := eval( y=F(x), ic ); solnC := solve( eq, {C} ); Y := unapply( eval( F(x), solnC ), x ); DEplot( diff(y(x),x) = f(x), y(x), x=0..2*Pi, [[y(Pi)=l]], color=black, linecolor=black, stepsize=0.05, title="Section 4.8 #103" ); Mathematica : (functions and values may vary) The following commands use the definite integral and the Fundamental Theorem of calculus to construct the solution of the initial value problems for exercises 103 - 105. Clear[x, y, yprime] yprime[x_] = Cos[x] 2 + Sin[x]; initxvalue = -it; inity value = 1; y[x_] = Integrate [yprime [t], {t, initxvalue, x}] + inityvalue If the solution satisfies the differential equation and initial condition, the following yield True yprime [x]==D[y[x],x] //Simplify y[initxvalue]==inity value Since exercise 106 is a second order differential equation, two integrations will be required. Clear[x, y, yprime] y2prime[x_] = 3 Exp[x/2] + 1; initxval = 0; inity val = 4; inityprimeval = — 1; yprime[x_] = Integrate[y2prime[t],{t, initxval, x}] + inityprimeval y[x_] = Integrate [yprime [t], {t, initxval, x}] + inityval Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative (blue). y2prime[x]==D[y[x], {x, 2}]//Simplify y[initxval]==inityval yprime[initxval]==inityprimeval Plot[{y[x], yprime[x]}, {x, initxval - 3, initxval + 3}, PlotStyle -> {RGBColor[l,0,0], RGBColor[0,0,l]}] CHAPTER 4 PRACTICE EXERCISES 1. No, since f(x) = x 3 + 2x + tan x => f'(x) = 3x 2 + 2 + sec 2 x > => f(x) is always increasing on its domain 2. No, since g(x) = esc x + 2 cot x =^ g'(x) = —esc x cot x — 2 esc 2 x =>- g(x) is always decreasing on its domain sin 2 x sin 2 x -\- (cos x + 2) < 3. No absolute minimum because lim (7 + x)(ll — 3x) 1,/3 = — oo. Nextf'(x) (11 -3x)^ -(7 + x)(ll -3xr 2 / 3 = (11 ( 7 1 3 l ) 3x y ) = ^$ Since f > if x < 1 and f < if x > 1, f(l) = 16 is the absolute maximum. ,T1 -3x)2/3 " i i ; _ 3x )2/3 => x = 1 and x = T are critical points. 4. f(x)=f±^ f'(x) a(x2 ' ( 1 ]:y +b) = " ( "^ + _ 2 y a) I f (3) = => -£ (9a + 6b + a) = => 5a + 3b = 0. We require also that f(3) = 1. Thus 1 3a+b 3a + b = 8. Solving both equations yields a = 6 and b = — 10. Now, f'(x) = 2(3x 2 _ 1)( * 3) so that f = | | +++ | +++ | . Thus f changes sign at x = 3 from (x 1} -1 1/3 1 3 positive to negative so there is a local maximum at x = 3 which has a value f(3) = 1. Cf igl (c) 1 Pearson Education, Inc., puiing as Pearson Addison-Wesle Chapter 4 Practice Exercises 275 5. Yes, because at each point of [0, 1) except x = 0, the function's value is a local minimum value as well as a local maximum value. At x = the function's value, 0, is not a local minimum value because each open interval around x = on the x-axis contains points to the left of where f equals — 1. 6. (a) The first derivative of the function f(x) = x 3 is zero at x = even though f has no local extreme value at x = 0. (b) Theorem 2 says only that if f is differentiable and f has a local extreme at x = c then f '(c) = 0. It does not assert the (false) reverse implication f'(c) = => f has a local extreme at x = c. 7. No, because the interval < x < 1 fails to be closed. The Extreme Value Theorem says that if the function is continuous throughout a finite closed interval a < x < b then the existence of absolute extrema is guaranteed on that interval. 8 . The absolute maximum is | — 1 1 = 1 and the absolute minimum is 1 1 = 0. This is not inconsistent with the Extreme Value Theorem for continuous functions, which says a continuous function on a closed interval attains its extreme values on that interval. The theorem says nothing about the behavior of a continuous function on an interval which is half open and half closed, such as [—1, 1), so there is nothing to contradict. 9. (a) There appear to be local minima at x = —1.75 and 1.8. Points of inflection are indicated at approximately x = and x = ± 1 . (b) f'(x) = x 7 - 3x 5 - 5x 4 + 15x 2 = x 2 (x 2 - 3) (x 3 - 5). The pattern y' indicates a local maximum at x = "y 5 and local minima at x = ± y 3 . (c) y = x 8 /8-x S /2-x 5 + 5x 3 f(x) = x 8 /8-x 6 /2-x 5 + 5x 3 I +++ I +++ I — | - -N/3 V^ ^ -+ 1.72 1.74 1.76 1.78 10. (a) The graph does not indicate any local extremum. Points of inflection are indicated at approximately x and x = 1. J . 600 / \ 400 / V"° / " 2 \ -201 / 2 -40C -60C f(x) -8" 5 - Sx (b) f'(x) = x 7 - 2x 4 - 5 + $ = x~ 3 (x 3 - 2) (x 7 - 5) . The pattern f )(+- • Sx-4r+ 11 X +++ indicates Copyright (c) 1 Pearson E Won, Inc., publishing as Pearson iii-ltt 276 Chapter 4 Applications of Derivatives a local maximum at x = y 5 and a local minimum at x = y2 . (c) 1 .07437 - 1 .2585 1.2599 11. (a) g(t) = sin 2 1 — 3t =$■ g'(t) = 2 sin t cos t — 3 = sin (2t) — 3 => g' < =>• g(t) is always falling and hence must decrease on every interval in its domain, (b) One, since sin 2 1 — 3t — 5 = and sin 2 1 — 3t = 5 have the same solutions: f(t) = sin 2 1 — 3t — 5 has the same derivative as g(t) in part (a) and is always decreasing with f(— 3) > and f(0) < 0. The Intermediate Value Theorem guarantees the continuous function f has a root in [—3, 0]. 12. (a) y = tan 8 =4> -=| = sec 2 6 > =>■ y = tan 9 is always rising on its domain =>• y = tan 6 increases on every interval in its domain (b) The interval [J, 7r] is not in the tangent's domain because tan is undefined at 8 = | . Thus the tangent need not increase on this interval. 13. (a) f(x) = x 4 + 2x 2 - 2 =>• f'(x) = 4x 3 + 4x. Since f(0) = -2 < 0, f(l) = 1 > and f'(x) > for < x < 1, we may conclude from the Intermediate Value Theorem that f(x) has exactly one solution when < x < 1 . (b) x 2 _ -2 ± V4 + 8 >0 => x 2 = a/3 - 1 and x > =4> x « v / - 7 320508076 « .8555996772 pyrj > 0, for all x in the domain of ^y =^ y = ^tj is increasing in every interval in 14. (a) y = ^ =► y' -- (x + its domain (b) y = x 3 + 2x => y' = 3x 2 + 2 > for all x => the graph of y = x 3 + 2x is always increasing and can never have a local maximum or minimum 15. Let V(t) represent the volume of the water in the reservoir at time t, in minutes, let V(0) = ao be the initial amount and V(1440) = ao + (1400)(43,560)(7.48) gallons be the amount of water contained in the reservoir after the rain, where 24 hr = 1440 min. Assume that V(t) is continuous on [0, 1440] and differentiable on (0, 1440). The Mean Value Theorem says that for some t in (0, 1440) we have V'(t ) = ^'^Ip^ = *o + (1400X43^60X7.48) - ao = 456 160,320 gal = 316778 1/min . Therefore at t the reservoir's volume 1440 1440 mm ' & u was increasing at a rate in excess of 225,000 gal/min. 16. Yes, all differentiable functions g(x) having 3 as a derivative differ by only a constant. Consequently, the difference 3x — g(x) is a constant K because g'(x) = 3 = -jj (3x). Thus g(x) = 3x + K, the same form as F(x). 17 Nq x = 1 ' ' ^ x ^ffi=--c fi«™ _ " ! x+l ' x+1 x+1 — x + d_ I x \ _ (x+l)-x(l) _ 1 &_ I -1 dx\x+l/ (x+1) 2 (x+1) 2 dxlx+1 j differs from — -j by the constant 1. Both functions have the same derivative 18. f'(x) = g'(x) 2x (x 2 +l) 2 f(x) — g(x) = C for some constant C =£> the graphs differ by a vertical shift. 19. The global minimum value of \ occurs at x = 2. Cop # (c) 1 Pearson Ediratio^, k, publishing as Pearson Addison-Wesle Chapter 4 Practice Exercises 277 20. (a) The function is increasing on the intervals [—3, —2] and [1, 2]. (b) The function is decreasing on the intervals [—2, 0) and (0, 1]. (c) The local maximum values occur only at x = —2, and at x = 2; local minimum values occur at x = —3 and at x = 1 provided f is continuous at x = 0. 21. (a) t = 0, 6, 12 (b) t = 3,9 (c) 6 < t < 12 (d) < t < 6, 12 < t < 14 (b) at no time (c) < t < 4 (d) 4<t< 24. y = x 3 -3x Z + 3 25. -x 3 + 6x 2 -9x + i 1 -1 \ 1 /2 3 4 -V \ 26. y=g-(x 3 + 3x~-9x-2? 27. 28. / i 500 400 y (6, 432) 300 200 /(4,256) \ 100 -2/^1 2 4 6 8l /-100 29. 30. ; y = x-3x 2 " i i. — , -3 i. 9 _J2-— " (8, -4) 27 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 278 Chapter 4 Applications of Derivatives 31. 32. 33. (a) y' = 16 — x 2 => y' = | +++ | => the curve is rising on (—4, 4), falling on (— oo, —4) and (4, oo) -4 4 => a local maximum at x = 4 and a local minimum at x = —4; y" = — 2x => y" -++| — the curve is concave up on (— oo, 0), concave down on (0, oo) =4> a point of inflection at x = (b) 34. (a) y' = x 2 - x - 6 = (x - 3)(x + 2) => y' = +++ | | +++ =>• the curve is rising on (-00, -2) and (3, 00), -2 3 falling on (—2, 3) =^ local maximum at x = —2 and a local minimum at x = 3; y" = 2x — 1 y" — ■ I +++ => concave up on (|, 00) , concave down on (—00, |) =>■ a point of inflection at x = | 1/2 (b) x=-2 x=1/2 x = 3 35. (a) y' = 6x(x + l)(x - 2) = 6x 3 - 6x 2 - 12x => y' the graph is rising on (—1, 0) — I +++I I+++ -10 2 and (2, 00), falling on (—00, — 1) and (0, 2) => a local maximum at x = 0, local minima at x = — 1 and x = 2; y" = 18x 2 - 12x - 12 = 6 (3x 2 - 2x - 2) = 6 (x - ±=fi\ (x - l - ± fi) +++ =>- the curve is concave up on I —00, 1— </7 1+V7 i-j/Z^ a„H (1±J1 and /7 A -,00 J concave down l-y/7 l + y/7 3^- ) =>• points of inflection at x = — y— (b) Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 36. (a) y' = x 2 (6 - 4x) = 6x 2 - 4x 3 ^ y' = +++ | +++ | 3/2 12x - 12x 2 = 12x(l - x) =>■ y" a local maximum at x = | ; y" Chapter 4 Practice Exercises 279 the curve is rising on (— oo, |), falling on (|, oo) - =4> concave up on I +++ I 1 (0, 1), concave down on (— oo, 0) and (1, oo) =>- points of inflection at x = and x = 1 (b) x = 37. (a) y' = x 4 - 2x 2 = x 2 (x 2 - 2) =4> y' = +- -y/2 V~2 +++ =>• the curve is rising on ( — oo, — v 2 ) and 2, oo 1 , falling on ( — y 2, y 2 J =$■ a local maximum at x = — y 2 and a local minimum at x = y 2 ; y" = 4x 3 - 4x = 4x(x - l)(x + 1) => y" = | +++ | | +++ => concave up on (-1, 0) and (1, oo), -1 1 concave down on (— oo, —1) and (0, 1) =>• points of inflection at x = and x = ± 1 (b) 38. (a) y' = 4x 2 - x 4 = x 2 (4 - x 2 ) => y' -c 4x(2-x 2 ) => y" +- the curve is rising on (—2, 0) and (0, 2), .+++1 — -2 2 falling on (— oo, —2) and (2, oo) => a local maximum at x = 2, a local minimum at x = —2; y" = 8x — 4x 3 concave up on ( — oo, — y 2 j and (0, y 2 J , concave -y/2 ° \[2 down on I — y 2, J and ( y 2, oo j =^> points of inflection at x = and x = ± y 2 -+ I — V~2 (b) x = 2 x = -2 39. The values of the first derivative indicate that the curve is rising on (0, oo) and falling on (— oo, 0). The slope of the curve approaches —oo as x — * CT, and approaches oo as x — » + and x — » 1. The curve should therefore have a cusp and local minimum at x = 0, and a vertical tangent at x = 1. Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 280 Chapter 4 Applications of Derivatives 2/3 , „>1/3 y = x + (x-1 ) 40. The values of the first derivative indicate that the curve is rising on (0, I) and (1, oo), and falling on (-co, 0) and (i, l) ■ The derivative changes from positive to negative at x = i indicating a local maximum there. The slope of the curve approaches —oo as x — » 0~ and x — * 1~, and approaches oo as x — » + and as x — > 1 + , indicating cusps and local minima at both x = and x = 1. 2/3 , . x 2/3 y = x + (x - 1 ) 41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches oo as x — > and as x — > 1, indicating vertical tangents at both x = and x = 1, y y' . 1 -2/3 1 / H v-2/3 3 2 1/3 , , ,1/3 y = x + (x - 1 ) i -3 -2 -1 ^ 1 2 3 -1 ^^^2 ■ """"* -3 42. The graph of the first derivative indicates that the curve is rising on ( 0, — z ^f — J and ( — ^- — , oo J , falling on (—oo, 0) and ( — z ^- — , — ^- — I =4> a local maximum at x = — z ^- — , a local minimum at x = — J j^- — . The derivative approaches — oo as x — > 0~ and x — > 1, and approaches oo as x — ► + , indicating a cusp and local minimum at x = and a vertical tangent at x = 1 . Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle Chapter 4 Practice Exercises y' 281 2/3 , H ,1/3 2 y = x - (x-1) -3 -2 1 2 3 43. y = |±1 = 1 4 x-3 t t+1 _i 1 44. y = ^_ = 2 - '" x + 5 x + 5 J\ y = 2 t 2X y_ X + 5 45. y x^ + 1 46. y x^ - X + 1 X- 1 Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 282 Chapter 4 Applications of Derivatives 47. y x J + 2 _ xf , 1 2x — 2 ' x 48. y x 4 - 1 _ „2 1 x 3 + 2 _ x* 1 2x 2 x 49. y = pEJ = 1 - ^ *=-vi 1 4 3 V 2 t x- 2 -3 j: = ^3 v=l -4 -3 1 -1 -1 -2 -3 1 /2 3 4 x =-2 x = 2 51. lim x^ 1 x 2 + 3x - 4 lim x-»l 2x + 3 1 52. lim x^ 1 x a -l x b -l lim x^ 1 bx b -i b 53. lim tanx = tan. = X — » 7T X 7T 54. lim lim x > n x + sinx x > q 1+ cos x 1 + 1 2 55. lim M% = lim 2 ™ x -, c ,°" = lim -^&, lim 2cos(2x) x tiro tan(x 2 ) x 1^ 2xsec2 ( x2 ) x^O 2xsec2 ( x2 ) x^O 2x(2sec 2 (x 2 )tan(x 2 )-2x) + 2sec 2 (x 2 ) + 2-1 56. lim ^"4 = lim "^"^ = s x y n sin(nx) x > n ncos(nx) n s(3x) x -> tt/2 - x -» tt/2 - lim -3sin(3x) _ 3 57. lim sec(7x)cos(3x) = lim . — mi. .. . ..- cos(7x) x— >7r/2~ _7sln ( 7x ) 7 58. lim ^/x sec x = lim .->— = - = ft x^0+ x^0 + V^ _ COS X 1 59. lim (esc x — cot x) = lim 1 — COS X x^O x^O lim x^ sin x Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Chapter 4 Practice Exercises 283 60. lim (4 - 4) = lim f 1 ^^ = lim (1 - x 2 ) • 4 = lim (1 - x 2 ) = lim 4 = 1 • oo = oo x ^0 Vx4 xV x^O V " 4 / x^O V ; x4 x^O V ; x^O x4 61. lim (Vx 2 + x+l- Vx 2 -x) = lim (Vx 2 + x + 1 - ^x 2 - xV ^ x2 + x + i+v^ lim 2x+ 1 X — > OO \A 2 + x + 1 + \/x 2 - x Notice that x = y x 2 for x > so this is equivalent to 2 + i lim X — > OO lim v^^+v^ x ^°° v /l + ' + i + v /rr ^ ^^ 62. lim x — » OO (_j! xM _ ,• x 3 (x 2 + 1) - xV - 1) _ l,x 2 -l ^ + l)~x^OO (x 2 -l)(x 2 + l) - x lim 2x 3 OO x * lim |4 -> oo ixi lim -» oo 12\2 12 lim x -» oo 24x lim ± = x ^ OO 2x 63. (a) Maximize f(x) = y/x- ^36 - x = x 1 / 2 - (36 - x) 1 / 2 where < x < 36 f'(x) -V 2 -U36-xr 1/2 (-l) y/36~- derivative fails to exist at and 36; f(0) = — 6, 2^ 2 W " "^ V ^ 2 v /x"\/36-x and f(36) = 6 => the numbers are and 36 (b) Maximize g(x) = ^fx + x/36 - x = x 1 / 2 + (36 - x) 1 / 2 where < x < 36 critical points at 0, 18 and 36; g(0) = 6, > g '(x)=Ix-V 2 + I(36-x)^(-l)=|^ 2 A '2 g(18) = 2VT8 = 6v^2 and g(36) = 6 => the numbers are 18 and 18 64. (a) Maximize f(x) = ^(20 - x) = 20X 1 / 2 - x 3 / 2 where < x < 20 => f'(x) = lOx" 1 / 2 - fx 1 / 2 _ 20 - 3x _ ^ x = and x = f are critical points; f(0) = f(20) = and f (f ) = Jf (20 - f ) 2\/x 3\/3 40 the numbers are y and , . (b) Maximize g(x) = x + a/20 - x = x + (20 - x) 1 / 2 where < x < 20 =^ g'(x) = ^LzJiz ' v/20" 79 x = y . The critical points are x = ™ and x = 20. Since g (™) = y and g(20) = 20, the numbers must be ™ and J . 65. A(x) = | (2x) (27 - x 2 ) for < x < \/ri => A'(x) = 3(3 + x)(3 - x) and A''(x) = -6x. The critical points are —3 and 3, but —3 is not in the domain. Since A"(3) = -18 < and A (y/zj) = 0, the maximum occurs at x = 3 => the largest area is A(3) = 54 sq units. ^-1. The x z v 2 , 128 66. The volume is V = x 2 h = 32 =>• h surface area is S(x) = x 2 + 4x (|f ) = where x > => S'(x) =4> the critical points are and 4, but is not in the domain. Now S"(4) = 2 + ^ > => atx = 4 there is a minimum. The dimensions 4 ft by 4 ft by 2 ft minimize the surface area. 2(x - 4) (x 2 + 4x + 16) x2 VIA Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 284 Chapter 4 Applications of Derivatives 67. From the diagram we have (|) + r 2 = ( y/3 I =>• r 2 = 12 ^ h . The volume of the cylinder is V = 7rr 2 h = n i 12 ^) h = f (12h - h 3 ) , where < h < 2 v^ ■ Then V'(h) = ^ (2 + h)(2 - h) =4> the critical points are —2 and 2, but —2 is not in the domain. At h = 2 there is a maximum since V' (2) = — 3ir < 0. The dimensions of the largest cylinder are radius = y/7. and height = 2. 68. From the diagram we have x = radius and y = height = 12 - 2x and V(x) = ± ttx 2 (12 - 2x), where < x < 6 =$> V'(x) = 2ttx(4 - x) and V'(4) = -8tt. The critical points are and 4; V(0) = V(6) = =4> x = 4 gives the maximum. Thus the values of r = 4 and h = 4 yield the largest volume for the smaller cone. Ah 69. The profit P = 2px + py = 2px + p ( 40 5 _ 1 ® X ) , where p is the profit on grade B tires and < x < 4. Thus P'(x) = (5 y. 2 (x 2 - lOx + 20) =>■ the critical points are [5 - V^J . 5, and [5 + y/Tj , but only n - y/S the domain. Now P'(x) > for < x < (5 - y/Sj and P'(x) < for (5 - y/yj <x<4=>atx=(5- is a local maximum. Also P(0) = 8p, P (5 - y/5j = 4p ( 5 - ^5) w 1 lp, and P(4) = 8p => at x = ( 5 - is an absolute maximum. The maximum occurs when x = ( 5 — y/ 5 ) and y = 2 ( 5 — y/ 5 ) , the units are hundreds of tires, i.e., x w 276 tires and y « 553 tires. 70. (a) The distance between the particles is |f(t)| where f(t) = —cos t + cos(t + |) . Then, f'(t) = sin t — sin(t Solving f'(t) = graphically, we obtain t « 1.178, t w 4.320, and so on. 2 (1.1780972.0 y/5\ there 5 ) there I)- Alternatively, f'(t) = may be solved analytically as follows, f (t) = sin (t + |) - | - sin (t + |) + | = ["sin (t + |) cos | -cos(t+ |)sinfl - [sin(t+ |)cosf + cos(t+ |) sin |1 = -2sin|cos(t+ |) so the critical points occur when cos(t + f ) = 0, or t = |+ lor. At each of these values, f(t) = ± cos ^ w ± 0.765 units, so the maximum distance between the particles is 0.765 units. Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle Chapter 4 Practice Exercises 285 (b) Solving cos t = cos (t + |) graphically, we obtain t ss 2.749, t w 5.890, and so on. Alternatively, this problem can be solved analytically as follows. cos t = COS ft + t) COS [(t+f) COS t n \ e,;^, TT cos(t+ IJcos I + sin(t + l)sin£ = cos(t+ £)cos| - sin(t+ £)sin 2sin(t + |) sin | = sin(t+|) =0 t The particles collide when t 2.749. (plus multiples of it if they keep going.) 71. The dimensions will be x in. by 10 - 2x in. by 16 - 2x in., so V(x) = x(10 - 2x)(16 - 2x) = 4x 3 - 52x 2 + 160x for < x < 5. Then V'(x) = 12x 2 - 104x + 160 = 4(x - 2)(3x - 20) , so the critical point in the correct domain is x = 2. This critical point corresponds to the maximum possible volume because V'(x) > for < x < 2 and V'(x) < for 2 < x < 5. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in. 3 Graphical support: 72. The length of the ladder is di + d2 = 8 sec 9 + 6 esc 9. We wish to maximize 1(9) = 8 sec 9 + 6 esc 9 => I' (9) = 8 sec 9 tan 9 - 6 esc 9 cot 9. Then I'(0) = 2 ^ sin 3 61-6 cos 3 9 = tan i di =4 36 and d 2 = 3 v / 36 J A + \/36 the length of the ladder is about (^4 + V36) \]a+ 3 v / 36 = (4 + V36) w 19.7 ft 73. g(x) = 3x-x 3 + 4 => g(2) Value Theorem. Then g'(x) = 3 so forth to x 5 = 2.195823345. 2>0andg(3) = -14 < 2 • 3x " 3x z > g(x) \ui 3-3xS in the interval [2, 3] by the Intermediate ; x = 2 => X! = 2.22 => x 2 = 2.196215, and Copyright (c) 2006 Pearson Education 286 Chapter 4 Applications of Derivatives 74. g(x) = x 4 - x 3 - 75 =>■ g(3) = -21 < and g(4) = 1 17 > =>• g(x) = in the interval [3, 4] by the Intermediate 3 T v 2 . v _ „ *n ~ *n ~ 75 Value Theorem. Then g'(x) = 4x 3 — 3x => x n+1 = x n =>■ x 2 = 3.229050, and so forth to x 5 = 3.22857729. ; x = 3 => xi = 3.259259 75. /(x 3 + 5x - 7) dx = £ + ^ - 7x + C 76. / (8t 3 - ^ + t ) di = f - $ + '- + C = 21 ' - i- + ^ + C 77. J(3 v / t+^)dt=/(3t 1 /2 +4 t- 2 )dt 3tV2 | 4t -l (I) , C = 2t 3 / 2 - 4 + C 78 - lfe-?) dt = /(K 1/2 -3t- 4 ) dt =H^ -^ + c = ^+? + c 79. Let u = r + 5 => du = dr J RV = J ^ = J u du= V + C = -11^+ C i +c (r + 5) + C 80. Let u = r - V2 =>■ du = dr /^^=6/^-L=6/^=6/u- 3 du = 6(^)+C = -3u- 2 + C -V^ 51. Let u = 6» 2 + 1 =4> du = 2(9 d6» => 1 du = 9 d6 J 36^6? + 1 dfl=/ A /5(|du) = |/ u 1 / 2 du = § ( s£ ) + C = u 3 /' 2 + C = (9 2 + if 2 + C 82. Let u = 7 + <9 2 => du = 26» d6» => | du = (9 d6> /7fe d0 = /7ua du ) = ^^ 1/2du =Hf +C = ul/2 + C=y7T ^ +C 83. Let u = 1 + x 4 => du = 4x 3 dx =^ | du = x 3 dx / x 3 (1 + x 4 r 1/4 dx = / U-V4 ( i du) = I J u-V4 du = I f^\ + C = \ u 3 / 4 + C = \ (1 + x 4 ) 3/4 + C 84. Let u = 2 - x =>• du = - dx => - du = dx J (2 - x) 3 / 5 dx = J u 3 / 5 (- du) = - j u 3 / 5 du • C = - | u 8 / 5 + C = - | (2 - x) 8 / 5 + C 85. Let u = ^ =>- du = i ds =>• 10 du = ds /see 2 ^ ds = J (sec 2 u) (10 du) = 10 J sec 2 u du = 10 tan u + C = 10 tan ^ + C 86. Let u = 7rs =>- du = n ds =>• - du = ds I esc 2 7ts ds = J (esc 2 u) (i du) = i J esc 2 u du = — i cot u + C = — ^ cot 7rs + C 87. Letu= v^f? =^ du= Jl&d => 4- du = d <? Jcsc v 7 ^ cot 1/20 d6» = J (esc u cot u) (4- du] = 4- (-esc u) + C = - 4- esc \fld + C Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Chapter 4 Additional and Advanced Exercises 287 Let u du = \ d9 => 3 du = dfl J sec | tan f d0 = / (sec u tan u)(3 du) = 3 sec u + C = 3 sec f + C Let u = | =4> du = j dx => 4 du = dx J sin 2 | dx = J (sin 2 u) (4 du) = J 4 ( '~ c 2 os2u ) du = 2/ (1 - cos 2u) du = 2 (u - ^) + C = 2u - sin 2u + C = 2 (|) - sin 2 (|) + C = § - sin | + C 90. Let u = | =4> du = i dx => 2 du = dx J cos 2 § dx = J (cos 2 u) (2 du) = J*2 ( 1+c 2 os2u ) du = / (1 + cos 2u) du = u + ^ + C f + \ sin x + C 91. y = J ^tl dx= J(l + x- 2 ) dx = x - x- 1 + C = x- ; + C; y = -1 when x = 1 =^> l-± + C = -l =► C=-l =► y = x-i-l 92. y = J (x + i) 2 dx = J (x 2 + 2 + 4j) dx = J (x 2 + 2 + x~ 2 ) dx = f + 2x - x" 1 + C = f + 2x - 1 + C; y = 1 when x=l => f+2-{+C=l => C y - 3 -f zx x 3 93. di / (l5v/t + X)dt= J (151 1 / 2 + 31- 1 / 2 ) dt = 10t 3 / 2 + 6t : / 2 + C; f = 8 when t = 1 => 10(1) 3 / 2 + 6(1) J / 2 + C = 8 => C = -8. Thus g = 10t 3 / 2 + 61 1 / 2 - 8 => r = J (lOt 3 / 2 + 6I 1 / 2 - 8) dt = 4t 5 /2 + 4t 3/2 _ 8t + C; r = when t = 1 => 4(1) 5 / 2 + 4(1) 3 / 2 - 8(1) + d = =>• Ci = 0. Therefore, r = 4t 5 / 2 + 4t 3 / 2 - 8t 94. |f = J -cos t dt = -sin t + C; r" = when t = =4> -sin + C = =4> C = 0. Thus, |f = -sin t => % = /-sintdt= cost + Ci;r' = whent = =>■ 1 + Ci = => Ci = -1. Then | = cos t - 1 =>• r = J (cos t - 1) dt = sin t - t + C 2 ; r = -1 when t = => - + C 2 = -1 =>■ C 2 = -1. Therefore, r = sin t — t — 1 CHAPTER 4 ADDITIONAL AND ADVANCED EXERCISES 1 . If M and m are the maximum and minimum values, respectively, then m < f(x) < M for all x e I. If m = M then f is constant on I. 2. No, the function f(x) 3x + 6, -2 < x < 9 - x 2 , < x < 2 has an absolute minimum value of at x = —2 and an absolute maximum value of 9 at x = 0, but it is discontinuous at x = 0. 3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical point. On a half-open interval the extreme values of a continuous function may be at a critical point or at the closed endpoint. Extreme values occur only where f ' = 0, f ' does not exist, or at the endpoints of the interval. Thus the extreme points will not be at the ends of an open interval. 4. The pattern ? = +++ minimum at x = 3. -++ indicates a local maximum at x = 1 and a local Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 288 Chapter 4 Applications of Derivatives 5. (a) If y' = 6(x+ l)(x - 2) 2 , theny' < Oforx < -1 and y' > Oforx > -1. The sign pattern is f = | +++ | +++ => fhas a local minimum at x = -1. Also y" = 6(x - 2) 2 + 12(x + l)(x - 2) -1 2 = 6(x — 2)(3x) =>■ y'' > for x < or x > 2, while y" < for < x < 2. Therefore fhas points of inflection at x = and x = 2. There is no local maximum, (b) If y' = 6x(x+ l)(x - 2), then y' < for x < -1 and < x < 2; y' > for -1 < x < and x > 2. The sign sign pattern is y' = — +++ | — | +++ . Therefore fhas a local maximum at x = and -1 local minima at x = -1 and x = 2. Also, y" = 18 |x - (^^Hl I* - ( ii 3^)l ' so f < ° 1—fl < x < H^lL and y" > for all other x => f has points of inflection at x = ^fl for 6. The Mean Value Theorem indicates that ^ijf* = f'(c) < 2 for some c in (0, 6). Then f(6) - f(0) < 12 indicates the most that f can increase is 12. 7. If f is continuous on [a, c) and f '(x) < on [a, c), then by the Mean Value Theorem for all x e [a, c) we have f "^lf° < => f(c) - f(x) < => f(x) > f(c). Also if f is continuous on (c, b] and f'(x) > on (c, b], then for all x G (c, b] we have f(x) : f(c) > =4> f(x) - f(c) > => f(x) > f(c). Therefore f(x) > f(c) for all x G [a, b]. (a) For all x, -(x + l) 2 < < (x - l) 2 => - (1 + x 2 ) < 2x < (1 + x 2 ) =» - \ < jf^ < \ . jf^ \<\, from part (a) (b) There exists c G (a, b) such that -^ = f( ^:f° f(b)-f(a) b-a |f(b) - f(a)| < i |b - a| 9. No. Corollary 1 requires that f '(x) = for all x in some interval I, not f '(x) = at a single point in I. 10. (a) h(x) = f(x)g(x) =>■ h'(x) = f'(x)g(x) + f(x)g'(x) which changes signs at x = a since f'(x), g'(x) > when x < a, f'(x), g'(x) < when x > a and f(x), g(x) > for all x. Therefore h(x) does have a local maximum at x = a. (b) No, let f(x) = g(x) = x 3 which have points of inflection at x = 0, but h(x) = x 6 has no point of inflection (it has a local minimum at x = 0). 11. From(ii), f(-l) b-c + 2 x+1 lim__f(x) = ^ lim__ bx2 + cx + 2 X — ► ± oo % x^±oo lim x^ ±oc x + c + a = 1; from (iii), either 1 = lim f(x) or 1 = lim f(x). In either case, lim r— ; — *r x^±oo bx + c + : and if c = 0, then lim X -> ± OO bx + 7 1 =$■ b = and c = 1. For if b = 1, then lim X — > ± 00 ± oo. Thus a = 1, b = 0, and c = 1. 12. | = 3x 2 + 2kx + 3 = => x k= ±3. -2k± y/4k 2 -36 6 x has only one value when 4k — 36 = => k = 9 or 13. The area of the AABC is A(x) = \ where < x < 1. Thus A'(x) (2)VT yr - X 2 = ( 1 — X and ± 1 are 2N1/2 critical points. Also A ( ± 1) = so A(0) = 1 is the maximum. When x = the AABC is isosceles since AC = BC= \/l. (x.Vl-x 2 )c Copffigl (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle Chapter 4 Additional and Advanced Exercises 289 14. lim f'(c + h)-f'(c) f"(c) =>• for e = \ |f"(c)| > there exists a 6 > such that < |h| < 6 f (c+h) - f'(c) - f"(c) < i |f"(c)| . Then f'(c) = £ |f"(c)| < f'(c + h) -f"(c)<i|f"(c)| => f"(c) - 1 |f"(c)| < ^^ < f"(c) + i |f"(c)| . If f"(c) < 0, then |f"(c)| = -f"(c) | f'( c ) < Li£±h) < i f '( C ) < 0; likewise if f "(c) > 0, then < \ f"(c) < ^£ + W < | f"( c ) h ^2 f '(c + h) < 0. Therefore, f(c) is a local (a) If f "(c) < 0, then -6 < h < => f '(c + h) > and < h < 6 maximum. (b) If f "(c) > 0, then -6 < h < => f'(c + h)< and < h < 6 =>■ f'(c + h) > 0. Therefore, f(c) is a local minimum. 15. The time it would take the water to hit the ground from height y is < / — , where g is the acceleration of gravity. The product of time and exit velocity (rate) yields the distance the water travels: D(y) v/64(h - y) = 8 J\ (hy - y 2 ) ' , < y < h => D'(y) = -4 J\ (hy - y 2 ) ' (h - 2y) => 0, | and h ,21-1/2, are critical points. Now D(0) = 0, D i 8h and D(h) = =>• the best place to drill the hole is at y 16. From the figure in the text, tan (/3 + 6) = ^±»; tan (/3 + 9) = ^^aZ*e \ and tan 9 = \ . These equations tan/3+j htan,fl + a give^ 1 tan/3 h — a tan Q Solving for tan /3 gives tan f3 bh h 2 + a(b + a) or (h 2 — a(b + a)) tan j3 = bh. Differentiating both sides with respect to h gives 2h tan P + (h 2 + a(b + a)) sec 2 (3 % = b. Then f = =► 2h tan /3 = b => 2h ( h2 + a b ( h b + a) ) 2bh 2 = bh 2 + ab(b + a) h 2 i(b + a) => h = v /a(a~+bj . 17. The surface area of the cylinder is S = 27rr 2 + 27rrh. From the diagram we have g = ^pp => h = RH R rIi and S(r) = 2?rr(r + h) = 2?rr (r + H - r g) = 2?r (l - |) r 2 + 27rHr, where < r < R. Case 1: H < R =^> S(r) is a quadratic equation containing the origin and concave upward =>■ S(r) is maximum at r = R. Case 2: H = R => S(r) is a linear equation containing the origin with a positive slope => S(r) is maximum at r = R. Case 3: H > R => S(r) is a quadratic equation containing the origin and concave downward. Then dS di we let r* 4tt (1 - |) r + 2ttH and f = => 4tt (l - | ) r + 2ttH = => r : RH 2(H-R) For simplification RH 2(H - R) (a) If R < H < 2R, then > H - 2R => H > 2(H - R) => 2 (h^r) > R which is im P ossible (b) IfH = 2R, thenr* 2R 2 2R R S(r) is maximum at r = R. (c) If H > 2R, then 2R + H < 2H => H < 2(H - R) H 2(H - R) < 1 => RH 2(H - R) <R r* < R. Therefore, RH S(r) is a maximum at r = r* = jt^ R) Conclusion : If H € (0, R] or H = 2R, then the maximum surface area is at r = R. If H e (R, 2R), then r > R which is not possible. If H 6 (2R, oo), then the maximum is at r = r* = 2(I RH R . 18. f(x) = mx - 1 + i => f'(x) = m - \ and f'(x) = \ > when x > 0. Then f'(x) = => x = -J- yields a XXX V IT1 l. If f ( -t= ) > 0, then A^/m — 1 + ^/m = 2^/m — 1 > => m>i. Thus the smallest acceptable value minimum. Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 290 Chapter 4 Applications of Derivatives for m is ^ . 2sin(5x) 19. (a) lim W x^O 3x lim 2sin(5x) X^O §( 5x ) X^O 3 ( 5x > lim &*°j& = m.i = i& (b) lim sin(5x)cot(3x) = lim sin(5x ' c °f x) W x^O K ' V ' x^O Sln < 3x ) lim ]• — 3sin(5x)sin(3x) + 5cos(5x)cos(3x) 5 x __j q 3cos(3x) — 3 (c) lim x esc V2x = lim — r 5 x -> x -» sin 2 ^ lim lim 2x x ^0 asm^a^^ x _, sin ^ 2 /2x") x^O cos(2v^x) 2_ = lim — t — =s— = X^O cos (2^/2x1 -2 (d) lim (sec x — tan x) X — > 7T/2 X — > 1 — COS x lim ±=^ -> tt/2 cos x lim =^ x^tt/2 - smx (e) lim ^f^ = lim v ' x j. n x — tan x x r\ 1 — sec^x lim x^O lim 55^ x -> tan x lim 9 - Slnx 2 x j, n 2 tan x sec^x lim ,, 2 s i n x cos^x = lim £2^ x^O ~ 2 (f) lim ^1 = x -> xsln x 2x cos(x 2 (g) x lim ^Jp 2x x^O ,. (x-2)(x 2 + 2x + 4) _ x lml 2 (x-2)(x + 2) -, -(2x 2 )sin(x 2 ) + 2cos(x 2 ) _ 2 — xsin x+2cos x 2 i* rn sec 3 x + tan 2 x sec x 1 + 1 lim ^xcus > x-; _ jj m -^x-^sunx-^i-zcos^A-; _ 2 _ ^ x > a xcos x+sin x x > n —xsin x+2cos x 2 ii rn sec x tan x ~ x^O lim x+2x + 4 — 4 + 4 + 4 -> 2 x+2 20. (a) lim yx + 5 V^+5 lim (b) lim 2x 6o x + 7,/x CO yA + 5 2x lim /1 + a X — > CO 1 ■ lim 2 CO x + 7 \/ x lim l ^°° 1+7, A 2 1+0 21. (a) The profit function is P(x) = (c — ex)x — (a + bx) = —ex 2 + (c — b)x — a. P'(x) = — 2ex + c — b = 2e . P"(x) = — 2e < Oif e > so that the profit function is maximized at x c-b 2e ' (b) The price therefore that corresponds to a production level yeilding a maximum profit is c — e 2e dollars. (c) The weekly profit at this production level is P(x) = -e(^) + (c - b)(- c_1> ] (d) The tax increases cost to the new profit function is F(x) = (c — ex)x — (a + bx + tx) = —ex Now F'(x) = -2ex + c - b - t = when x = ^p = ' L ^ 1 - Since F"(x) = -2e < if e > 0, F is maximized a=^-a. 2 -J- <c - b - t)x - a. when x 2e units per week. Thus the price per unit is p = c — e( c 2 ^ ' ) c + b + t dollars. Thus, such a tax increases the cost per unit by c+ ^ +t — £ ^ = | dollars if units are priced to maximize profit. 22. (a) The x-intercept occurs when i — 3 = 0=>^=3=^x=i. (b) By Newton's method, x n+1 = x n - jS|4-. Here f'(x n ) = -x~ 2 = ^. So x n+1 = x n = x n + x n ox n = ZX n ox n = X n ^Z oX n J. Xn + --3Vn Copffigl (c| 1 Pearson link, fc, publishing as Pearson Addison-Wesle Chapter 4 Additional and Advanced Exercises 291 f(*o) f'(*o) X o~ a _ qxg - xg - a _ xg(q-l)-a i x 3 q^" 1 q^" 1 — | — — Xq ( - — ) + -5=1 I - J so that xi is a weighted average of Xq and -^=j with weights mo q-i and nil = - . q q In the case where x„ = ^we have x q = a and Xl = ^ (*=i) + -^ (l) = -^ (a_i + l) a x o 24. We have that (x - h) z + (y - h) z = r 2 and so 2(x - h) + 2(y - h)g = and 2 + 2^ + 2(y - h)3J = hold >dy ,<fy rfi ,dy Thus 2x + 2y2i = 2h + 2hf, by the former. Solving for h, we obtain h x + y — jjr ■ Substituting this into the second equation yields 2 + 2^ + 2y ax 2 *±y£ 0. Dividing by 2 results in 1 dy i v d!z dx T J dx 2 *+yg 25. (a) a(t) = s"(t) = -k (k > 0) => s'(t) = -kt + Cj, where s'(0) Ci = 88 => s'(t) -kt s(t) -kt 2 2 -kt 2 2 C 2 where s(0) = =>■ C 2 = so s(t) -kt 2 ., + 88t. Now s(t) = 100 when + 88t = 100. Solving for t we obtain t = 88 ± ^f ~ 200k . At such t we want s'(t) = 0, thus ' 88 + y^ 2 - 200k " _ K / of -h y oo~ — ZUUK \ Oor kf 88 -^ 8 f- 200k ) So so that k = |§g w 38.72 ft/sec 2 . (b) The initial condition that s'(0) = 44 ft/sec implies that s'(t) 0. In either case we obtain ! -kt 2 5 2 200k = -kt + 44 and s(t) = -jp + 44t where k is as above. The car is stopped at a time t such that s'(t) = — kt + 44 = => t = y . At this time the car has traveled a distance < i lt) = f(f) +44(f) = f stopping distance. k 968 (fj§) = 25 feet. Thus halving the initial velocity quarters 26. h(x) = f 2 (x) + g 2 (x) => h'(x) = 2f(x)f'(x) + 2g(x)g'(x) = 2[f(x)f'(x) + g(x)g'(x)] = 2[f(x)g(x) + g(x)(-f(x))] = 2-0 = 0. Thus h(x) = c, a constant. Since h(0) = 5, h(x) = 5 for all x in the domain of h. Thus h(10) = 5. 27. Yes. The curve y = x satisfies all three conditions since -=| = 1 everywhere, when x = 0, y = 0, and -^ everywhere. 28. y' = 3x 2 + 2 for all x =>■ y = x 3 + 2x + C where -l = l 3 + 2-l + C=>C = -4^y = x 3 + 2x-4. 29. s"(t) = a = -t 2 =>• v = s'(t) C. We seek Vq = s'(0) = C. We know that s(t*) = b for some t* and s is at a maximum for this t*. Since s(t) = -£ + Ct + k and s(0) = we have that s(t) = -£ + Ct and also s'(t*) = so that t* = (3C) 1/3 - [-(3C) 1 / 3 ] 4 So C(3C) 1/3 ubr Thus v = s'(0) (4b) J b^(3C) 1/3 (C-§) 12 (3C) 1/3 (f)=b^3 1 /3 C 4 / 3 =f 30. (a) s"(t) = t 1 / 2 - r 1 ! 2 =>■ v(t) = s'(t) = ft 3 / 2 - 2I 1 / 2 + k where v(0) 4 K ~ 3 J_t5/2 _ l t 3/2 15 l 3 l v ( t ) = | t 3/2 _ 2t l/2 + 4 3 l 15 ' (b) s(t) = j|t 5 / 2 - ft 3 / 2 + ft + k 2 where s(0) = k 2 = -±. Thus s(t) 31. The graph of f(x) = ax 2 + bx + c with a > is a parabola opening upwards. Thus f(x) > for all x if f(x) = for at most one real value of x. The solutions to f (x) = are, by the quadratic equation (2b) 2 - 4ac < => b 2 - ac < 0. :yW~' Thus we require 32. (a) Clearly f(x) = (aix +bi) + . . . + (a n x + b n ) > for all x. Expanding we see f(x) = (a 2 x 2 + 2 ai b lX + b 2 ) + . . . + (a 2 x 2 + 2a n b n x + b 2 ) = (a 2 + a 2 + . . . + a 2 )x 2 + 2(aib a + a 2 b 2 + . . . + a n b n )x + (b 2 + b Thus (aibj + a 2 b 2 + . . . + a n b n ) 2 - (a 2 + a 2 + . . . + a 2 )(b 2 + b 2 + • • + b 2 ) > 0. b 2 ) < by Exercise 31. Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 292 Chapter 4 Applications of Derivatives Thus (aibi + a 2 b 2 + . . . + a n b n ) 2 < (af + a| + . . . + ajj)(b? + b 2 + . . . + bjj). (b) Referring to Exercise 31: It is clear that f (x) = for some real x O b 2 — 4ac = 0, by quadratic formula. Now notice that this implies that f(x) = (aix + b : ) 2 + . . . + (a„x + b n ) 2 = (a 2 + a 2 + . . <^> (aibi + a 2 b 2 + <^> (aibi + a 2 b 2 + 2(aibi + a 2 b : 2°2 a n b n )x + (b 2 bl b) + a n b n ) 2 - (a 2 + a 2 + . . . + a 2 )(b 2 + b ar,b n (a 2 + a 2 + ... + a 2 )(b 2 +b 2 +.. But now f(x) = •£>• a,x + b; = for all i = 1, 2, . . . , n •«• a;x ■ -bi + b 2 ) = + b 2 ) = for all i 1,2, n. Copyright (c) 2006 Pearson Education CHAPTERS INTEGRATION 5.1 ESTIMATING WITH FINITE SUMS (a) Ax = ^ = \ and X; = iAx (b) Ax = i^ = \ and x; = iAx Since f is increasing on [0, 1], we use left endpoints to obtain lower sums and right endpoints to obtain upper sums. a lower sum is E(i) , -l = l(o , + (l) , ) = i (c) Ax = iyS = i and x; = iAx = | =^> an upper sum is $^(5) ' " = " ' ^ 4 ' i=l i-n alowersumisE(t) 2 -| = U° 2+ (3) 2+ (5) 2 +(I) 2 ) i=o v ' 2) ' 2 ~~ 2 \ V2V T± / — 8 I I 4 ' 8 J7_ ■ VI (d) Ax i-o _ 1 I and x; = iAx = \ =4> an upper sum is ^ ( l _ l / 1\ :d 2 +(!) 2 ^ '30^ v 16 J 15 ■ VI Since f is increasing on [0, 1], we use left endpoints to obtain lower sums and right endpoints to obtain upper sums. (a) Ax = Ljp = \ and x; = iAx = \ a lower sum is (b ) Ax=i^ = i (c) Ax = iffl = 1 and x } = iAx = | (d) Ax = i=o = i and x . = iAx = i Ea)'4=i(o"+(i)')=^ i=0 v y andx, = iAx = \ ^ a lower sum is £(i) 3 ■ \ = | (o 3 + (i) 3 + Q) 3 + (f) 3 ) i=o v 7 uppersumisX:(i) 3 -i = |(G) 3 +1 3 ) = £ • | = & -g(i) 3 -! = i(a) 3 +(D 3 +(i) 3 +i 3 ) an upper sum i 36 __ 9 256 64 100 _ 25 256 64 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 294 Chapter 5 Integration 3. f(x) Since f is decreasing on [0, 1], we use left endpoints to obtain upper sums and right endpoints to obtain lower sums. (a) Ax = £f± = 2 and xi = 1 + iAx = 1 + 2i =>■ a lower sum is £ j: ' 2 = 2 (| + £ i-1 4 15 (b) Ax = ^j 1 = 1 and xi = 1 + iAx = 1 + i =>- a lower sum is^F' 1 = 1 (5 + 5 + 3 + 5 i=l 1 i 1 i 1 i 1> _ II 60 (c) Ax 5-1 2 and x; = 1 + iAx = 1 + 2i => an upper sum is ^^ • 2 = 2(l + 4 i=0 3 (d) Ax = ^-j 1 = 1 and x, = 1 + iAx = 1 + i =>• an upper sum is^F' 1 = 1 ( 1 +5 + 5 + 3) = ff i=0 4. f(x) = 4 - x 2 Since f is increasing on [—2, 0] and decreasing on [0, 2], we use left endpoints on [—2, 0] and right endpoints on [0, 2] to obtain lower sums and use right endpoints on [—2. 0] and left endpoints on [0, 2] to obtain upper sums. (a) Ax = 2 [ 2) = 2 and x ; = -2 + iAx = -2 + 2i =>• a lower sum is 2 • (4 - (-2) 2 ) + 2 • (4 - 2 2 ) = 1 4 (b) Ax = ^t^ = 1 and x ; = -2 + iAx = -2 + i => a lower sum is ]£ (4 - (x ; ) 2 ) • 1 + ]T (4 - (x ; ) 2 ) • 1 i=0 i=3 = 1((4 - (-2) 2 ) + (4 - (-1) 2 ) + (4 - l 2 ) + (4 - 2 2 )) = 6 (c) Ax = 2 ~^~ 2) = 2 and x ; = -2 + iAx = -2 + 2i => a upper sum is 2 • (4 - (0) 2 ) + 2 • (4 - 2 ) = 16 (d) Ax 2 -(-2) 1 and x; = -2 + iAx = -2 + i => a upper sum is £ (4 - (x ; ) 2 ) • 1 + ]T (4 - (x ; ) 2 ) • 1 i=l 3 — r i-2 1((4 - (-1) 2 ) + (4 - 2 ) + (4 - 2 ) + (4 - l 2 )) = 14 Using 2 rectangles =>• Ax ¥ = }=>iWi)m)) l((i) 2 + (!) 2 ) 10 _ _5_ 32 16 1-0 _ 1 4 4 Using 4 rectangles => Ax =>i(f(l)+f(|) + f(|)+f(|)) = 4-(Q) 2 + (f) 2 + (I) 2 + (f) 2 ) = ! Copy fight (c| 1 Pearson Etati, he, publishing as Pearson Addison-Wesle Section 5.1 Estimating with Finite Sums 295 Using 2 rectangles ^Ax = 1-0 _ 2 1 2 = 2-((3) 3 + (D 3 \ 28 ) 2-64 7 32 Using 4 rectangles => Ax = 1-0 _ 4 1 4 =►*(*(*) +f(§) +f(l)+f(D) 1 ( l 3 +3 3 +5 3 +7 3 A ~ 4 \ 8 3 j 496 ~~ 4-8 3 ~~ 124 8 3 — 31 12.x l(f(i)+f(!)) 7. f(x) = i y f(x) = 4 - x 2 —i 1 — ►- x 4 5 Jsing 2 rectangles =>• Ax = 5^1 =2^2 (f(2) + f(4)) = 2(| + 1) = | Jsing 4 rectangles => Ax = ^ = 1 ^l(f(l)+f(l)+f(I) +f(D) = 1( 2 + 2 + 2 + 2-) = 1488 _ 496 _ ■5-7-9 5-7-9 496 315 Jsing 2 rectangles => Ax - 2 - { - 2) -2^2(f( -l) + f(l)) = 2(3 + 3) = 12 Jsing 4 rectangles => Ax _ 2- (-2) _ 1 4 - 1 - =M(f(-|)+f(-|)+f Q)+f(D) = l((4-(-l) 2 ) + ( 4 -H) 2 ) + ( 4 ~{\ )V(4- = 16 -(| -2 + 1-2) = 16 -f = 11 ! )) 9. (a) D « (0)(1) + (12)(1) + (22)(1) + (10)(1) + (5)(1) + (13)(1) + (H)(1) + (6)(1) + (2)(1) + (6)(1) = 87 inches (b) D « (12)(1) + (22)(1) + (10)(1) + (5)(1) + (B)(1) + (H)(1) + (6)(1) + (2)(1) + (6)(1) + (0)(1) = 87 inches 10. (a) D « (1)(300) + (1.2)(300) + (1.7)(300) + (2.0)(300) + (1.8)(300) + (1.6)(300) + (1.4)(300) + (1.2)(300) + (1.0)(300) + (1.8)(300) + (1.5)(300) + (1.2)(300) = 5220 meters (NOTE: 5 minutes = 300 seconds) (b) D « (1.2)(300) + (1.7)(300) + (2.0)(300) + (1.8)(300) + (1.6)(300) + (1.4)(300) + (1.2)(300) + (1.0)(300) + (1.8)(300) + (1.5)(300) + (1.2)(300) + (0)(300) = 4920 meters (NOTE: 5 minutes = 300 seconds) 11. (a) D « (0)(10) + (44)(10) + (15)(10) + (35)(10) + (30)(10) + (44)(10) + (35)(10) + (15)(10) + (22)(10) + (35)(10) + (44)(10) + (30)(10) = 3490 feet « 0.66 miles (b) D « (44)(10) + (15)(10) + (35)(10) + (30)(10) + (44)(10) + (35)(10) + (15)(10) + (22)(10) + (35)(10) + (44)(10) + (30)(10) + (35)(10) = 3840 feet « 0.73 miles 12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the midpoints of each time interval to approximate this area using rectangles. Thus, D « (20)(0.001) + (50)(0.001) + (72)(0.001) + (90)(0.001) + (102)(0.001) + (112)(0.001) + (120)(0.001) + (128)(0.001) + (134)(0.001) + (139)(0.001) « 0.967 miles (b) Roughly, after 0.0063 hours, the car would have gone 0.484 miles, where 0.0060 hours = 22.7 sec. At 22.7 sec, the velocity was approximately 120 mi/hr. Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 296 Chapter 5 Integration 13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing acceleration • At. Thus, At = 1 and speed w [32.00 + 19.41 + 11.77 + 7.14 + 4.33](1) = 74.65 ft/sec (b) Using right end-points we obtain a lower estimate: speed rs [19.41 + 11.77 + 7.14 + 4.33 + 2.63](1) = 45.28 ft/sec (c) Upper estimates for the speed at each second are: t 1 2 3 4 5 V 32.00 51.41 63.18 70.32 74.65 Thus, the distance fallen when t = 3 seconds is s w [32.00 + 51.41 + 63.18](1) = 146.59 ft. 14. (a) The speed is a decreasing function of time attained. Also right end-points give an lower estimate for the height (distance) t 1 2 3 4 5 V 400 368 336 304 272 240 gives the time-velocity table by subtracting the constant g = 32 from the speed at each time increment At = 1 sec. Thus, the speed w 240 ft/sec after 5 seconds, (b) A lower estimate for height attained is h w [368 + 336 + 304 + 272 + 240](1) = 1520 ft. 15. Partition [0,2] into the four subintervals [0,0.5], [0.5, 1], [1, 1.5], and [1.5,2]. The midpoints of these subintervals are mi = 0.25, iri2 = 0.75, ni3 = 1.25, and mi = 1.75. The heights of the four approximating rectangles are f( mi ) = (0.25) 3 = ^ f(m 2 ) = (0.75) 3 = g, f(m 3 ) = (1.25) 3 = if, and f(m 4 ) = (1.75) 3 - ^ Notice that the average value is approximated by \ :d'(§; ;!)'(§; " 64 64 :i) 3 (i: 31 16 length of [0,2] approximate area under curve f(x) = x 3 We use this observation in solving the next several exercises. 16. Partition [1 , 9] into the four subintervals [1, 3], [3, 5], [5, 7], and [7, 9]. The midpoints of these subintervals are mi = 2, ni2 = 4, 1113 = 6, and mi = 8. The heights of the four approximating rectangles are f(mi) = \, f(m2) = t, f(ni3) = |, and f(iri4) = =, The width of each rectangle is Ax = 2. Thus, Area 2 a) 2( )+2Q; 25 12 average value : area length of [1,9] (I) 8 25 96" 17. Partition [0,2] into the four subintervals [0,0.5], [0.5, 1], [1, 1.5], and [1.5,2]. The midpoints of the subintervals are mi = 0.25, m 2 = 0.75, m 3 = 1.25, and m 4 = 1.75. The heights of the four approximating rectangles are 1 c ;„2 5ir _ 1 f(mi) 1 I _* 2 7T tj + sin -r l,f(m 2 ) \ + sin 2 2f l,f(m 3 ) k} v^ = j + ^ = 1, and f(m4) — , Area«(l + l + l + l)(i) =2 1+sin2 T = l + (-^)^ average value length of [0,2] 1 . The width of each rectangle is Ax 1 = 1. i Thus, 18. Partition [0, 4] into the four subintervals [0, 1], [1,2, ], [2, 3], and [3,4]. The midpoints of the subintervals are mi = | , m 2 = | , m 3 = | , and m 4 = | . The heights of the four approximating rectangles are f(mi) = 1 - (cos (^ir)) = l ~ ( cos (I)) 4 = 0-27145 (to 5 decimal places), f(m 2 ) = 1 - (cos (^)) = 1 - (cos (f )) 4 = 0.97855, f(m 3 ) = 1 - (cos (^)) = 1 - (cos (f )) 4 = 0.97855, and f(m 4 ) = 1 - (cos (^)) = 1 - (cos(^)) 4 = 0.27145. The width of each rectangle is Ax=l. Thus, Area « (0.27145)(1) + (0.97855)(1) + (0.97855)(1) + (0.27145)(1) = 2.5 => average value area length of [0,4] 225 _ 5 4 — 8' Copyright (c) 2006 Pearson Education Section 5.1 Estimating with Finite Sums 297 19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left endpoints: (a) upper estimate = (70)(1) + (97)(1) + (136)(1) + (190)(1) + (265)(1) = 758 gal, lower estimate = (50)(1) + (70)(1) + (97)(1) + (136)(1) + (190)(1) = 543 gal. (b) upper estimate = (70 + 97 + 136 + 190 + 265 + 369 + 516 + 720) = 2363 gal, lower estimate = (50 + 70 + 97 + 136 + 190 + 265 + 369 + 516) = 1693 gal. (c) worst case: 2363 + 720t = 25,000 =>■ t«31.4hrs; best case: 1693 + 720t = 25,000 =>■ t w 32.4 hrs 20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate uses left endpoints: (a) upper estimate = (0.2)(30) + (0.25)(30) + (0.27)(30) + (0.34)(30) + (0.45)(30) + (0.52)(30) = 60.9 tons lower estimate = (0.05)(30) + (0.2)(30) + (0.25)(30) + (0.27)(30) + (0.34)(30) + (0.45)(30) = 46.8 tons (b) Using the lower (best case) estimate: 46.8 + (0.52)(30) + (0.63)(30) + (0.70)(30) + (0.81)(30) = 126.6 tons, so near the end of September 125 tons of pollutants will have been released. 21. (a) The diagonal of the square has length 2, so the side length is y 2. Area = (b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring 2tt TV 16 — 8' Area = 16fJ) (sin |) (cos |) = 4 sin f = 2 a/2 « 2.828 (c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring 2tt _ jr_ 32 16 - Area = 32(i) (sin i) (cos ^] sin 2^2 w 3.061 (d) Each area is less than the area of the circle, n. As n increases, the area approaches ir. 22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring || = |. The area of each isosceles triangle is At = 2(g) (sin |) (cos ^) = | sin ^. (b) The area of the polygon is Ap = nAx = § sin — , so lim § sin — = lim tt ■ -j^f- = n 1 n n^oo^ n n^oo \T) (c) Multiply each area by r 2 . A T = ir 2 sin — 1 2 n A P = 2r 2 sin 2l r 2 n limAp = 7rr 2 n— >oo -26. Example CAS commands: Maple: with( Student[Calculusl] ); f := x -> sin(x); a:=0; b := Pi; plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); N:=[100, 200, 1000]; for n in N do Xlist := [ a+l.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: for n in N do Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); #(b) #(c) Copyright (c) 2006 Pearson Education 298 Chapter 5 Integration end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); FunctionAverage(f(x),x=a..b,output=plot); # (d) fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); Mathematica : (assigned function and values for a and b may vary): Symbols for n, — > , powers, roots, fractions, etc. are available in Palettes (under File). Never insert a space between the name of a function and its argument. Clear[x] f[x_]:=x Sin[l/x] {a,b} = {7r/4, tt} Plot[f[x],{x, a, b}] The following code computes the value of the function for each interval midpoint and then finds the average. Each sequence of commands for a different value of n (number of subdivisions) should be placed in a separate cell. n=100;dx = (b-a)/n; values = Table[N[f[x]], {x, a + dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n n =200; dx = (b - a) /n; values = Table[N[f[x]],{x, a + dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n n=1000;dx = (b-a)/n; values = Table[N[f[x]],{x, a + dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]j] / n FindRoot[f[x] == average, {x, a}] 5.2 SIGMA NOTATION AND LIMITS OF FINITE SUMS 1- E E 6k 6(1) , 6(2) _ 6 , 12 + 1 1 + 1 """ 2+1 2^3 z " ^ k 1 T 2 ^ 3 u ^ 2 ^ 3 6 3. J2 cos kit = cos(l7r) + cos(27r) + cos(37r) + cos(47r) = —1 + 1 — 1 + 1=0 k=l 4. J2 sin k7r = sin (l7r) + sin (27r) + sin (37r) + sin (4tt) + sin (57r) = + + + + = 5. £ (-l) k+1 sin I = (-1) 1+1 sin f + (~1) 2+1 sin f + (-1) 3+1 sin f = - 1 + ^ = A^ 6. Y, (-1)" cos kn = (-1) 1 cos(Itt) + (-1) 2 cos(2tt) + (-1) 3 cos(3tt) + (-1) 4 cos(4?r) k=l = -(-!)+ !-(-!)+ 1 = 4 Cop # (c) 1 Pearson Ediratio^, k, publishing as Pearson Addison-Wesle Section 5.2 Sigma Notation and Limits of Finite Sums -' - 2 1 - 1 + 2 2 - 1 + 2 3 - 1 + 2 4 - 1 + 2 5 - 1 + 2 6 - 1 (b) ± 2 k = 2° + 2 1 + 2 2 + 2 3 + 2 4 + 2 5 = 1 + 2 + 4 + (c) E 2 k ^ 299 7. (a) E 2"- 1 =2' ' : 2 k=l 1 + 2 + 4 + 8+16 + 32 + 16 + 32 2 -i 1 + 2 o+i + 2 1+1 + 2 2+1 + 2 3+1 + 2 4+1 = 1+2 + 4+8+16 + 32 All of them represent 1+2 + 4 + 8+16 + 32 (a) E (-2) k -' = (-2) 1 - 1 + (-2) 2 - 1 + (-2) 3 - 1 + (-2) 4 - 1 + (-2) 5 - 1 + (-2) 6 - 1 = 1-2 + 4-8+16-32 k=l (b) J2 (-l) k 2 k = (-l)°2° + (-l) 1 2 1 + (-l) 2 2 2 + (-l) 3 2 3 + (-l) 4 2 4 + (-l) 5 2 5 =1-2 + 4-8+16-32 k=() (c) J2 (-l) k+1 2 k + 2 = (-l)- 2+1 2- 2 + 2 + (-1)- 1+1 2- 1+2 + (-1) 0+1 2 0+2 + (-1) 1+1 2 1+2 + (-1) 2+1 2 2 + 2 k=-2 + (-l) 3+1 2 3 + 2 = -1 +2-4 + 8- 16 + 32; (a) and (b) represent 1 — 2 + 4 — 8+16 — 32; (c) is not equivalent to the other two 9- (a) E ^l + (-1)* 2-1 ^ 3-1 1 + 2 3 k+1 0+l'l+l'2+l 2'3 (b)E^ = ^ + ^-^ (c) E k + 2 -1 + 2 ' + 2 ' 1+2 + •1 + i - i (a) and (c) are equivalent; (b) is not equivalent to the other two. 10. (a) E ( k - !) 2 = (1 - I) 2 + (2 - l) 2 + (3 - l) 2 + (4 - l) 2 = + 1 + 4 + 9 k=l (b) E (k + I) 2 = (-1 + I) 2 + (0 + l) 2 + (1 + l) 2 + (2 + l) 2 + (3 + l) 2 = + 1 + 4 + 9 + 16 k=-l (c) E k 2 = (-3) 2 + (-2) 2 + (-l) 2 = 9 + 4+l k=-3 (a) and (c) are equivalent to each other; (b) is not equivalent to the other two. 11. Ek 12. E k 2 13. E i 14. E 2k 15. E(-D k+1 i 16. E(-D k I 17. (a) E 3a k = 3 E a k = 3(-5) = -15 k=l k=l (b) E h i = ltw = l(6)=l k=l k=l (c) E (a k + b t ) = E a t + E b t = -5 + 6 = 1 k=l k=l k=l (d) E (a k - b t ) = E a t - E bk = -5 - 6 = -11 k=[ k=l (e) E (bk - 2a k ) = Eb k -2Ea k = 6- 2(-5) = 16 k=l k=l k=l Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 300 Chapter 5 Integration 18. (a) £ 8a k = 8 E a k = 8(0) = n n (c) E (a* + 1) = E a K + E 1 = ° + n = n k=l k=l k=l (b) E 2 50b k = 250 E W = 250(1) = 250 k=l n n (d) E(b k -D = Eb k -El = l-n k=l k=l k=l 19. (a) Ek 10(10+1) 55 (c) £ k 3 = [wQO+iij 2 = 552 = 3025 (b) Ek : 2 _ 10(10+ 1)(2(10)+1) 385 20. (a) Ek k=l 13 13(13-1) 2 91 (c) J2 k 3 = [i32|±i2l = 91 s 8281 (b) Ek 2 _ 13(13 + 1X2(13)+ 1) 819 21. E -2k = -2 E k = -2 f 71 ^ 12 ) -56 22. E f5 = BEk=f3^)=- k=l k=l x ' 1 , _ V^ 1 V^ v2 _ ^^^ 6(6 + l)(2(6)+l) 23. E(3-k 2 )=E3-Ek 2 = 3(6) -73 k=l k=l A <\ - SP V2 V^ c _ 6(6+l)(2(6)+l) 24. £( k 2-5) = Ek 2 -E5 5(6) = 61 25. E k(3k + 5) = E (3k 2 + 5k) = 3 E k 2 + 5 E k = 3 ( 5(5 + 1)( 6 2(5)+1) ) + 5 f 5 ^) V — 1 1- — 1 V — 1 V — 1 ** / \ / 240 k=l k=l 26. E k(2k + 1) = E (2k 2 + k) = 2Ek 2 + Ek = 2 ( 7(7 + 1)( 6 2(7)+1) ) 1,-1 1.— 1 1-— i i — i V * 7(7+1X2(7)+!)^ , 7(7+1) 308 k=l k=l 27. g ^+(Ek) 3 = ^Ek3 + (E k y=^(^ 2 +(5^) 3 = 3376 k=l / k=l 28. Ek -Ef= Ek -iEk 3 = (^XH 4 ^ =588 29. (a) (b) (c) (2,3) 0<x<2 Left-hand c, =0 c ? c/= 1 c d 2 a- 2 L 3 , M (2,3) /(x)=x z -l, < .t < 2 Right-hand Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Section 5.2 Sigma Notation and Limits of Finite Sums 301 30. (a) (b) (c) f(x)=-x z -1 f(x) = -x' "■" crm f(x)=-x ^J 31. (a) (b) (c) f(x) = sin x, -ir<X<TT Right-hand h XTX. 32. (a) 2 1.5 Willi -Jt -K/2 -0.5 -1 :■■:■:■:■:/.■;:.■;-:■:■:■■;. (b) 71/2 it f(x) = sin x +1 y 2r .M.w. . .M ., 1.5 -jt -ji/2 -0.5 -1 (c) Jl/2 It f(x) = sin x +1 2 1 f 1.5 / 111 viy -Jt -rc/2 -0.5 -1 Jt/2 7C f(x) - sin x +1 33. |xj - x | = |1.2- 0| = 1.2, |x 2 - xi| = |1.5 - 1.2| = 0.3, |x 3 - x 2 | = |2.3 - 1.5| = 0.8, |x 4 - x 3 | = |2.6 - 2.3 1 = 0.3, and |x 5 - x 4 | = |3 - 2.6| = 0.4; the largest is ||P|| = 1.2. 34. | Xl - x | = 1-1.6 - (-2)| = 0.4, |x 2 - xj| = | — 0.5 - (-1.6)| = 1.1, |x 3 - x 2 | = |0 - (-0.5)| = 0.5, |x 4 -x 3 | = |0.8-0| =0.8, and|x 5 - x 4 | = |1 - 0.8| = 0.2; the largest is ||P|| = 1.1. Since f is decreasing on [ 0, 1] we use left endpoints to obtain upper sums. Ax = ^— ^ = - and x; = iAx = -. So an upper sum is E(i-*?)H^S(i-(;o 2 ) = ^£(n 2 -i 2 ) i=0 i=0 v ' i=0 ^ 2 = i i=0 i=0 x ' i=0 (n-l)n(2(n-l) + l) _ 1 2n 3 -3n 2 +n 6n 3 1 -^ — — . Thus, 6 ' lim E(l-xf)i=Mm 1- n— >oo .ft n— t™ v l - Gn3 12 3 3 Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 302 Chapter 5 Integration Since f is increasing on [ 0, 3] we use right endpoints to obtain upper sums. Ax = ^2 = 2 anc j Xj = [/\ x = 21 Soan U pp er sum is E2x 5 (|)= Ef -I = ^Ei i=l i=l i=l 18 n(n + l) _ 9n 2 + 9n n 2 2 ~~ n 2 Thus, lim E~ • " = lim^ 9n 2 + 9n ,2 lim (9- n— >oo 9. Since f is increasing on [ 0, 3] we use right endpoints to obtain upper sums. Ax = ^2 = 3 an( j x . _ j^ x _ & Soan upper sumisE(x? + l)l= E((!) 2 + l)! = |E(f + l i=l i=l v x i=l v ' = fEi 2 + f-n=g( n(n + 1) c (2n + 1) )+ 3 i-l 9(2n 3 + 3n 2 + n) 2n 3 o 18 +f + 5 , o Tk 3 = 5 — =- + 3. Thus, lim E(x 2 + l)i=lim 18+^ + - 3 =9 + 3 = 12. Since f is increasing on [ 0, 1] we use right endpoints to obtain upper sums. Ax = -!— 2 — I and x; = iAx = L . Soan upper sum is X>? (I) = E3(^) 2 (D = I Ei 2 = Jr • ( n(n+ T n + 1) ) i=l i=l i=l v ' 2n J + 3n 2 + n • '-•■" 2 ti_2. Thus, lim E3x, 2 (i) 2 n^oo g ' Kn) 2n 3 = lim n^oo 2+S + 2 j 2 ± - 39. f(x) = x + x 2 =x(l + x) y Since f is increasing on [ 0, 1] we use right endpoints to obtain upper sums. Ax = ^— ^ — I anc [ Xj — j/\ x — I So an upper sum i=l i=l v i=l i=l 2n 3 + 3n 2 + n 1 / n(n+l) \ 1 / n(n+l)(2n + l) \ _ n 2 + n , n 2 \^ 2 J + n l\ 6 y~ 2n 2 "+" n Thus, lim E( x i + x2 )s 6n 3 i±i 2 3,1 = lim n^oo m n^oo 2 + 2 + 4, 6 1 , 2 _ 5 2 ' G G - Since f is increasing on [ 0, 1] we use right endpoints to obtain upper sums. Ax = -!— 2 — I and x; = iAx = -. Soan upper sum isE(3 X i + 2x 2 )i = E(f + 2(D 2 )n = ,li:i+4Ei 2 i=l i=l v • i=l i=l 3n 2 + 3n , 2n 2 + 3n + 1 3 f n(n+l) \ 2 f n{n+ l){2n + 1) \ n 2 V 2 ) + n 3 ^ - 6 " ) 2n 2 3n 2 2 + ; + -V -+ 3 " 2 . Thus, lim E(3x, + 2x 2 )i = lim n^oo m n^oo 2 + n + S 3 3 j. 2 _ 13 2^3 - Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 5.3 THE DEFINITE INTEGRAL Section 5.3 The Definite Integral 303 1. fx 2 dx Jo 4. f 1 dx J 1 x r° 7. | (sec x) dx J -tt/4 2. dx 8. J (tan x) dx h i-x 3. J\(x 2 -3x)dx 6. J a/4 - x 2 dx 9. (a (c (c (f) 10. (a (b (c (d (e (f) 11. (a (c 12. (a (c 13. (a (b 14. (a (b / g(x)dx = (b) f 5 g(x) dx = -f i g(x) dx = -8 J 3f(x)dx = 3j f(x) dx = 3(-4) = -12 (d) J f(x) dx = J f(x) dx - J f(x) dx = 6 - (-4) = 10 / x [f(x) - g(x)] dx = £ f(x) dx - J x g(x) dx = 6 - 8 = -2 Jj [4f(x) - g(x)] dx = 4 Jj f(x) dx - f t g(x) dx = 4(6) -8 = 16 n9 p9 J -2f(x)dx = -2J f(x)dx= -2(-l) = 2 n9 p9 ^9 J. [f(x) + h(x)] dx = J f(x) dx + J. h(x) dx = 5 + 4 = 9 J 7 [2f(x) - 3h(x)] dx = 2 j 7 f(x) dx - 3 J 7 h(x) dx = 2(5) - 3(4) = -2 f g f(x) dx = - J] f(x) dx = -(-1) = 1 / x f(x) dx = J] f(x) dx - J 7 f(x) dx = -1 - 5 = -6 J* g [h(x) - f(x)] dx = J 7 [f(x) - h(x)] dx = j 7 f(x) dx - f 7 h(x) dx = 5 - 4 = 1 J] f(u) du = J] f(x) dx = 5 j 2 if « dt = - r f « dt = - 5 (b) J x y/3 f(z) dz = a/3 / x f(z) dz = 5 a/3 (d) J* [-f(x)] dx = - Jj f(x) dx = -5 /; g (t)dt=-/_ 3 g(t)dt=-A/2 f_ a [-g(x)] dx = - j 3 g(x) dx = -a/2 / a f(z) dz = f o f(z) dz - J fl f(z) dz = 7 - 3 = 4 J 4 3 f(t)dt=-/ 3 4 f(t)dt=-4 (b) J 3 g(u) du = J_ a g(t) dt = a/2 »> r 3 f dr =v3r3^ dt =fe)(^)= 1 p3 n3 r« 1 J h(r) dr = J h(r) dr - J h(r) dr = 6 - = 6 - J 3 h(u) du = - ( - J] h(u) du") = Jj h(u) du = 6 Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 304 Chapter 5 Integration 15. The area of the trapezoid is A = | (B + b)h = \ (5 + 2)(6) = 21 => J 2 (| +3) dx = 21 square units 16. The area of the trapezoid is A = I (B + b)h P 3/2 = i (3 + !)(!) = 2 f 3/2 J (-2x + 4)dx 1/2 2 square units 17. The area of the semicircle is A 2 " L 2 i-(3) 2 /-3 ^ - ^ dx 7r square units 18. The graph of the quarter circle is A |^(4) 2 47r =>• I v 1 6 — x 2 dx = 47r square units ■f f(x) - -2x + 4 0,5 1 1,5 2 -3 -2 -1 12 3 19. The area of the triangle on the left is A = \ bh = \ (2)(2) = 2. The area of the triangle on the right is A = \ bh |(1)(D Then, the total area is 2.5 r |x| dx = 2.5 square units 20. The area of the triangle is A = \ bh = \ (2)(1) = 1 =>• I (1 — |x|) dx = 1 square unit Section 5.3 The Definite Integral 305 y f(x) = 1-|x| 21. The area of the triangular peak is A = \ bh = \ (2)(1) = 1. The area of the rectangular base is S = £w = (2)(1) = 2. Then the total area is 3 => I (2 — jx|) dx = 3 square units 22. y = 1 + \f\ -X 2 => y -\ = y/\ -X 2 =>■ (y - l) 2 = 1 - x 2 =>• x 2 + (y - l) 2 = 1, a circle with center (0, 1) and radius of 1 => y = 1 + y 1 — x 2 is the upper semicircle. The area of this semicircle is A = 1 7rr 2 = 1 7r(l) 2 = |. The area of the rectangular base is A = £w = (2)(1) = 2. Then the total area is 2 + | I ll + vl— x 2 )dx = 2+f square units ) = 1 +Vi-> 23. J b fd^=5(b)(^) 1 /wt^ _ b^ 4 r b 24. J 4x dx = \ b(4b) = 2b 2 4b iV" 4 " Copffigl (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 306 Chapter 5 Integration 25. J 2s ds = \ b(2b) - \ a(2a) = b 2 - a 2 y-2s 26. J a 3t dt = \ b(3b) - \ a(3a) = § (b 2 - a 2 ) / 3b A) / i 3a / i i i i ■ a b ,y = a 27. f^d^-f = | /; J 1 29. 5 d0 (2tt) 2 _ ^ 3tt1 2 2 2 3,//WM H 33. J'l/2 ,^3 t 2 dt=if = i J. 2.") 0.5 28 . | 2 ; 5 xdx= (2|)!_(0|)! = 3 B ^_M! Ml 2 30. J,, rdr 2 2 24 X 0.3 32. | s 2 ds = &P- = 0.009 /W 2 m 3 t 34. 6> 2 d6» — — - Jo 3 — 24 J. 2a a 35. f xd x=^-f = f 37. ^b X dx 39. J 7dx = 7(l-3) = -14 41. J o 5xdx = 5 J o xdx = 5 [ 2 2 2 10 43. 44. / o 2 (2t-3)dt = 2j r i 1 tdt-/ o 2 3dt = 2[ L 36. I xdx=£$>--£=.» r 3b X 2 Jo (3b) 3 38. | x 2 dx = ^ = 9b 3 'o 3 40. J. -2 dx= V2(-2-0)= -2\/2 42 - fl\**=\fl***=\[\ 1 | 51 31 _ 16 _ i 2 2 16 - 1 21 _ o! 2 2 3(2 - 0) = 4 - 6 = -2 r t - V2 dt p\/2 P\/2 _ Jo tdt -Jo ^ dt fV£ -A/2U/2-0 = 1 -2= -1 45. |„ (1 + 1) dz 46. / 2 1 ld Z+ / 2 1 ldz=/ 2 1 ldz-l/ i 2 zdz=l[l-2]-l[f-f] 2 V2/< 4 nO nO nO f.3 pO f , J 3 (2z - 3) dz = J 3 2z dz - J 3 3 dz = -2 J q z dz - J 3 3 dz = -2 If - f - 3[0 - 3] = -9 + 9 = 47. /;3u 2 du = 3/>du = 3[£u 2 du-/\ 2 du]=3([f-f]-[f-f])=3[f-^=3(|)=7 Copy |t (c| 1 Pearson Etati, Inc., publishing as Pearson Addison-Wesle Section 5.3 The Definite Integral 307 /, i pi 2 f, U2 J 1/2 24u 2 du = 24 | u 2 du = 24 /2 J 1/2 pi pl/2 I u 2 du — I u Jo Jo du 24 24 (I) r<3 x- + x - 5) dx = 3 / x 2 dx / x 2 dx + / x dx I Jo Jo Jo xdx-| 5 dx = 3 I ^ - ^ o Jo f - f - 5[2 - 0] = (8 + 2) - 10 = 50. J] (3x 2 + x - 5) dx = - J o (3x 2 + x - 5) dx = - 3 J Q x 2 dx + J Q x dx - J q 5 dx 0r-¥) + 0r-S)-5(i-o)]=-(i-5) = ! 51. Let Ax = ^ = ^ and let x = 0, Xi = Ax, n n u ' i ' X2 = 2Ax, . . . , x n _j = (n — l)Ax, x n = nAx = b. Let the c k 's be the right end-points of the subintervals => ci = xi, C2 = X2, and so on. The rectangles defined have areas: f(ci) Ax = f(Ax) Ax = 3(Ax) 2 Ax = 3(Ax) 3 f(c 2 ) Ax = f(2Ax) Ax = 3(2Ax) 2 Ax = 3(2) 2 (Ax) 3 f(c 3 ) Ax = f(3Ax) Ax = 3(3Ax) 2 Ax = 3(3) 2 (Ax) 3 f(c n ) Ax = f(nAx) Ax = 3(nAx) 2 Ax = 3(n) 2 (Ax) 3 Then S n = ]T f(c k ) Ax = £ 3k 2 (Ax) 3 3(Ax) 3 J2 k k= k=l 3V 1-2-3 b 3 2 + L f\?\ / n(n+l)(2n+l) \ r Jo 3x 2 dx = lim £(2+^ + 4, n ^ oo 2 V ' n ' n 2 b 3 . 3b f (x) = 3x (b,3b Xo-0 x, x 2 -x„_, x„.b 52. Let Ax = ^_J> = k and let x = 0, Xi = Ax, X2 = 2Ax, . . . , x n _j = (n — l)Ax, x n = nAx = b. Let the c k 's be the right end-points of the subintervals =^> ci = xi, C2 = X2, and so on. The rectangles defined have areas: f(ci) Ax = f(Ax) Ax = tt(Ax) 2 Ax = tt(Ax) 3 f(c 2 ) Ax = f(2Ax) Ax = tt(2Ax) 2 Ax = tt(2) 2 (Ax) 3 f(c 3 ) Ax = f(3Ax) Ax = tt(3Ax) 2 Ax = tt(3) 2 (Ax) 3 f(c n ) Ax = f(nAx) Ax = Tr(nAx) 2 Ax = 7r(n) 2 (Ax) 3 Then S n = £ ffe) Ax = ]T 7rk 2 (Ax) 3 k=l k=l = *(&*)'£ k' = ir(g)( » + l f + 11 ) Jo 7Tb 3 n + n^ Tb :; 7tx 2 dx = lim n — » oc o '2+! + A) Ttb V 2 (b.Jtb ) 3 • f(x) = JtX 2 / / -"' X x. -b Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 308 Chapter 5 Integration 53. Let Ax = ^ = b and let = Q = A n n u ' -l X2 = 2Ax, . . . , x n _! = (n — l)Ax, x n = nAx = b. Let the c k 's be the right end-points of the subintervals => Ci = xi, C2 = X2, and so on. The rectangles defined have areas: f(ci) Ax = f(Ax) Ax = 2(Ax)(Ax) = 2(Ax) 2 f(c 2 ) Ax = f(2Ax) Ax = 2(2Ax)(Ax) = 2(2)(Ax) 2 f(c 3 ) Ax = f(3Ax) Ax = 2(3Ax)(Ax) = 2(3)(Ax) 2 f(c n ) Ax = f(nAx) Ax = 2(nAx)(Ax) = 2(n)(Ax) 2 Then S n = t %0 Ax = £ 2k(Ax) 2 2(Ax) 2 Ek = 2(|)( n ^l) b 2 1 k=l 1 Jo 2x dx lim b 2 1 + n — > oo v 2b v° 1 f(x) = 2x / / / 54. Let Ax = ^ = ^ and let x = 0, Xi = Ax, n n u ' i ) X2 = 2Ax, . . . , x n _j = (n — l)Ax, x n = nAx = b. Let the c k 's be the right end-points of the subintervals =^> ci = xi, C2 = X2, and so on. The rectangles defined have areas: f(ci) Ax = f(Ax) Ax = (^ + l) (Ax) = \ (Ax) 2 + Ax f(c 2 ) Ax = f(2Ax) Ax = (2^ + l) (Ax) = ^(2)(Ax) 2 + Ax f(c 3 ) Ax = f(3Ax) Ax = (2j* + l) (Ax) = \ (3)(Ax) 2 + Ax f(c) Ax = f(nAx) Ax = (sf* + l) (Ax) = \ (n)(Ax) 2 + Ax Then S n = t ffe) Ax = £ (± k(Ax) 2 + Ax) = \ (Ax) 2 tk + Axfl k=l k=l k=l k=l 1 (b 2 2 [ n^ ib 2 o + ^: 1 dx lim i b 2 1 n — > oo \4 \ b) |b 2 + b. n(n+l) (n) • Vs 55. av(f) = (-^) J o \x 2 -l)dx x 2 dx - 4- I 1 dx \/3 Jo ^ - (A = i - i = o. i v^ 56. av(f)=(3^)/ o 3 (-f)dx=H-DX 3 _ I f 3£^ - _3._x^_ _ 3 6 I 3 J 2 ' 2 2 ' x 2 dx Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 57. av(f) = (^5) J o (-3x 2 - 1) dx = = - 3 / lx2dx -X lidx =- 3 (C)- (1 - 0) Section 5.3 The Definite Integral 309 58. av(f»=( T ^)/ o 1 (3x 2 -3) dx X lx2dx -X l3dx=3 (^)- 3(1 - 0) 59. av(f)=(3^)/ o 3 (t-l) 2 dt n3 p3 r*6 = Uo t2dt -IJo tdt +Uo 1 = i(?)-i(f-f) + |(3-o> = i- dt 0.5 1 1.5 2 15 3 60. av(f)=( T ^)/_ 1 2 (t 2 -t) dt i/_>dt-i£tdt 3 \tJ — 3 V - 3~ J ~*~ 2 3 2 2 -2 -1.5 -1 -0.5 61. - l)dx (a) av(g) = (jj-s) j\ (|x| - 1) = i/_° i (-x-l)dx+i/ o 1 (x-l)dx /0 nO nl nl -i« to -U-i ldx +Uo xdx -Uo 1 dx l(f-^)-|(o-(-i)) + |(f-f)-|a-o) .. g(x)-M-i -1 H — k—» ■+—J 1 1 X Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 310 Chapter 5 Integration (b) av(g) = (3^) J x (|x| - 1) dx = \ Jj (x - 1) dx = iX 3 xdx-ij; 3 ldx=l(f-|)-i(3-l) (c) av(g) = (arfzij) /^(M - 1) dx = i/ 1 i (|x|-l)dx+i/ i 3 (|x|-l) dx 7 (— 1 + 2) = i (see parts (a) and (b) above). H 1 1 H -!•■ ■■ y-M-1 ■3 • ■2 ■ 1 -t 1 X 12 3 ••2 H h -1 y-M-1 12 3 < 1 X 62. (a) av(h) = (0^) J_ r |x| dx = f_ r (-x) dx x dx £ _ (-1) 2 _ _ 1 2 2 2 ■ (b) av(h) = to)/o 1 -l« (V _ &\ _ _ 1 ^ 2 2 ^ 2' dx ■/.' Jo x dx (c) av(h) = (^) £ - |x| c ' /.O pi J-f-Mfc + Jo-l* = 2 (~ 2 + (~ 2)) = ~ 2 ( see P^ 8 W and (b) above). •■ 1 -t — 1 — 1 — t- h(x)=-|x| -I 1 1 1 X ■■-1 ■ ■ 1 -i 1 1 H h(x) = -|x| H 1 1 X •■-1 ■■ 1 ■4 1 ► h(x)=-|x| I 1 1 1 X ••-1 63. To find where x - x 2 > 0, let x - x 2 = and b = 1 maximize the integral. => x(l - x) = => x = Oorx = 1. If < x < 1, then < x - x 2 =^ a = Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 5.3 The Definite Integral 311 64. To find where x 4 - 2x 2 < 0, let x 4 - 2x 2 = => x 2 (x 2 - 2) = =>• x = or x = ± \fl. By the sign graph, ++++++ +++++++, we can see that x 4 - 2x 2 < on l-y/l, \0\ =>a=-v/2andb V^ minimize the integral. 65. f(x) = yt-3 is decreasing on [0, 1] =>■ maximum value of f occurs at =£• max f = f(0) = 1 ; minimum value of f occurs at 1 => min f = f(l) = jr^s = \ ■ Therefore, (1—0) min f < I jr-^ dx < (1 — 0) max f => k < I y-Lj dx < 1 . That is, an upper bound = 1 and a lower bound 66. See Exercise 65 above. On [0, 0.5], max f = y+& = 1. m in f = fxfea = 0.8. Therefore (0.5 - 0) min f < J Q f(x) dx < (0.5 - 0) max f => \ < J Q -^ dx < \ . On [0.5, 1], max f = l+ \ 05)2 = 0.8 and f = -r-^2 = 0.5. Therefore (1 - 0.5) min f < \ r-jU dx < (1 - 0.5) max f =>- \ < f y-f-j dx l + V v ' — Jo.5 l+ x — 4 — Jo.5 '+ x ~n c ^ i _ i mm Then \ + 1 < 5 ' < n0.5 pi nl I T-U dx + I r-i-, dx < ± + f- =>• ^ < I -r-^ dx < JO 1 + x J 0.5 1 + x — 2 ' 5 20 — J o 1 + x — 9_ 10 ' 67. -1 < sin(x 2 ) < 1 for all x => (1 -0)(-l) < J sin(x 2 )dx < (1 - 0)(1) or J sin x 2 dx < 1 =4> J equal 2. sin x 2 dx cannot 68. f(x) = v/x + 8 is increasing on [0, 1] => max f = f(l) = y/i + 8 = 3 and min f = f(0) = v/O + 8 = 2\/l . Therefore, (1 - 0) min f < \ Jx + 8 dx < (1 - 0) max f => 2a/2 < f \/x + 8 dx < 3. *J t/ 69. If f(x) > on [a, b], then min f > and max f > on [a, b]. Now, (b — a) min f < / f(x) dx < (b — a) max f. Then b > a => b - a > => (b - a) min f > => J f(x) dx > 0. 70. If f(x) < on [a, b], then min f < and max f < 0. Now, (b - a) min f < J f(x) dx < (b - a) max f. Then b > a =4> b - a > =>• (b-a) max f < => J f(x) dx < 0. 71. sin x < x for x > =4> sin x — x < for x > =>■ I (sin x — x) dx < (see Exercise 70) =>■ I sin x dx — I x < =4> I sin x dx < I x dx =>• I sin x dx < ( y — y J =4> I sin x dx < \ . Thus an upper bound is \ . dx 72. sec x > 1 + \ on 2 \ 2 ' 2 / sec x — (l + j)> 0cm (-!,!) =► /^[secx-^ + f^dx^O Exercise 69) since [0, 1] is contained in (— |, |) => I sec x dx — I (l + y)dx>0=> I sec x dx (see > £( 1 + l9 dx =* X SeCxdx ^£ ldx+ 2/ x2dx => X secxdx >( 1 -0)+l(T) =* J secxdx>|. Thus a lower bound is I. 73. Yes, for the following reasons: av(f) = ^-^ I f(x) dx is a constant K. Thus / av(f) dx = I K dx r*b pb r»h = K(b - a) =4> J av(f) dx = (b - a)K = (b - a) • ^ J f(x) dx = J f(x) dx. Cop # (c) 1 Pearson Ediratio^, Inc., publishing as Pearson Addison-Wesle 312 Chapter 5 Integration 74. All three rules hold. The reasons: On any interval [a, b] on which f and g are integrable, we have: (a) av(f + g) = j^_ J a [f(x) + g(x)] dx = ^ | J] f(x) dx + X g(x) dx = ^-J & f(x) dx + ^ J] g(x) dx = av(f) + av(g) (b) av(kf) = g^ X kf W dx = b^i [ k J a f « dx ] = k [shi X f(x) dx = k av(f) r»b pb nb (c) av(f) = ^ J a f(x) dx < ^L J a g(x) dx since f(x) < g(x) on [a,b], and ^A J^ g(x) dx = av(g). Therefore, av(f) < av(g). 75. Consider the partition P that subdivides the interval [a, b] into n subintervals of width Ax = ^^ and let c^ be the right endpoint of each subinterval. So the partition is P = {a, a + ^=^, a + ~ , . . . , a + ~ a) } and Ck = a H — ~ . We get the Riemann sum ^f(c k )Ax = JTc ■ ^ = ^^E 1 = £ ^ r il ■ n = c(b - a). As n -» oo and ||P|| -> k=l k=l ' k=l b this expression remains c(b — a). Thus, I c dx = c(b — a) 76. Consider the partition P that subdivides the interval [a, b] into n subintervals of width Ax = - — - and let c^ be the right endpoint of each subinterval. So the partition is P = {a, a + ^F- 2 , a H — ( ~ , . . ., a + "^ ~ *' } and c^ = a + ~ ■ We get the Riemann sum £f(c k ) Ax = L>k(^) = l ^E(a+ k ^) 2 = ^E (a 2 + b-a f V^ Q 2 i 2a(b-a) ^ E^+^pEk+^Ek 1 k ! k=l k=l x ' k=l 2ak(b - a) k 2 (b-a)' k-1 k-1 k-1 (b-a^y^^l _ b-a „„2 i 2a(b-a)^ n(n+l) , (b-a) J n(n+l)(2n+l) n na n- + ii- ii (b-a)a 2 + a(b-a) 2 .i±i + <V l! -^4 ±i = (b - a)a 2 + a(b - a) 2 • *±I + ^ • ™^' - "• ^~ 2 As n — » oo and ||P|| — > this expression has value (b — a)a 2 + a(b — a) • 1 + * ~ a) • 2 = ba 2 - a 3 + ab 2 - 2a 2 b + a 3 + i(b 3 - 3b 2 a + 3ba 2 - a 3 ) = f - f . Thus, J a x 2 dx = f - f . 77. (a) U = maxi Ax + max2 Ax + . . . + max n Ax where maxi = f(xi), max2 = f(x2), . . . , max n = f(x n ) since f is increasing on [a, b]; L = mini Ax + min2 Ax + . . . + min n Ax where mini = f(xo), min2 = f(xi), . . . , min n = f(x n _i) since f is increasing on [a, b]. Therefore U — L = (maxi — mini) Ax + (max2 — min2) Ax + . . . + (max„ — min„) Ax = (f(xi) - f(x )) Ax + (f(x 2 ) - f(xi))Ax + . . . + (f(x n ) - f(x n _ { )) Ax = (f(x n ) - f(x )) Ax = (f(b) - f(a)) Ax. (b) U = maxi Axi -I- max 2 Ax2 + . . . + max„ Ax n where maxi = f(xi), max2 = ffe), . . . , max n = f(x n ) since f is increasing on[a, b]; L = mini Axi + min2 Ax 2 + . . . + min„ Ax„ where mini = f(xo), min 2 = f(xi), . . . , min n = f(x n _i) since f is increasing on [a, b]. Therefore U — L = (maxi — mini) Axi + (max2 — min2) Ax2 + . . . + (max n — min n ) Ax n = (f(xi) - f(x )) Axi + (f(x 2 ) - f(xj))Ax 2 + . . . + (f(x n ) - f(x n _,)) Ax„ < (f(xi) - f(x )) Ax raM + (f(x 2 ) - f( Xl )) Ax max + . . . + (f(x n ) - f(x n _,)) Ax max . Then U - L < (f(x n ) - f(x )) Ax max = (f(b) - f(a)) Ax max = |f(b) - f(a)| Ax max since f(b) > f(a). Thus lim (U - L) = lim (f(b) - f(a)) Ax mK = 0, since Ax max = ||P|| . ||P|| -► ||P|| - " " Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 5.3 The Definite Integral 313 + max n Ax where max n = f(x n _!) +• min n Ax where . min n = f(x n ) 78. (a) U = maxi Ax + max2 Ax + . maxi = f(xo), max2 = f(xi), . since f is decreasing on [a, b]; L = mini Ax + min2 Ax + . . mini = f(xx), min 2 = f(x 2 ), . . since f is decreasing on [a, b]. Therefore U — L = (maxi — mini) Ax + (max 2 — min 2 ) Ax + . . . + (max n — min n ) Ax = (f(x ) - f( Xl )) Ax + (f( Xl ) - f(x 2 ))Ax + . . . + (ffr^) - f(x n )) Ax = (f(x ) - f(x n )) Ax = (f(a)-f(b))Ax. (b) U = maxi Axi + max 2 Ax 2 + . . . + max,, Ax n where maxi = f(xo), max 2 = f(xi), . . . , max n = ftx,,^) since f is decreasing on[a, b]; L = mini Ax x + min 2 Ax 2 + . . . + min,, Ax„ where mini = f( x i), min 2 = f(x 2 ), . . . , min n = f(x n ) since f is decreasing on [a, b]. Therefore U — L = (maxi — mini) Axi + (max 2 — min 2 ) Ax 2 + . . . + (max n — min„) Ax n = (f(x ) - f(xi)) Axi + (f(xi) - f(x 2 ))Ax 2 + . . . + (fCx^) - f(x„)) Ax n < (f(x ) - f(x n )) Ax mM = (f(a) - f(b) Ax max = |f(b) - f(a)| Ax,„ M since f(b) < f(a). Thus 0, since Ax m „ = ||P|| . lim (U - L) ||P||-0 lim o |f(b) - f(a)| Ax„ 79. (a) Partition [0, |] into n subintervals, each of length Ax = |j with points Xo = 0, Xi = Ax, x 2 = 2 Ax, ... , x„ = nAx = |. Since sin x is increasing on [0, |] , the upper sum U is the sum of the areas of the circumscribed rectangles of areas f(xi) Ax = (sin Ax)Ax, f(x 2 ) Ax = (sin 2Ax) Ax, . . . , f(x„) Ax (sin nAx) Ax. Then U = (sin Ax + sin 2Ax sin nAx) Ax cos 4f -cos( (n+ 1) Ax) 2 sin ss Ax COS^- COS (( n+ l) jL) 2 sin | 4] n/2 7r(cos^-cos(| + ^)) _ COS X_ cos (| + jL) (b) The area is I sin x dx r Jo lim cos f — cos 4n sin -f- 4n (f + fe) 4n / m 80. (a) The area of the shaded region is J2^ x i ' m i which is equal to L. i=l n (b) The area of the shaded region is X]^ x i ' Mj which is equal to U. i=l (c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part of the figure and the first part of the figure. Thus this area is U — L. n n 81. By Exercise 80, U — L = J]^ x i ' Mj — J]^ x i ' m i where Mj = max{f(x) on the ith subinterval} and i=i i=i n n mi = min{f(x) on the ith subinterval}. Thus U — L = ^fMj — m;)Ax; < J^e ■ Axj provided Ax; < 6 for each i=l i=l n n i = 1, . . . , n. Since J2 e ' ^ x i = e S^ x i = e 3 ~ a ) m e result, U — L < e(b — a) follows. i=l i=l Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 314 Chapter 5 Integration 82. The car drove the first 150 miles in 5 hours and the second 150 miles in 3 hours, which means it drove 300 miles in 8 hours, for an average of ^9 mi/hr = 37.5 mi/hr. In terms of average values of functions, the function whose average value we seek is < t < 5 5< 1 < (30)(5) + (50)(3) = 37 5 Velocity mi/hr v 501 V(t) f 30, \50, 30 and the average value is _average "value 37.5 mi/hr -t Time hr 83-88. Example CAS commands: Maple : with( plots ); with( Student[Calculusl] ); f :=x -> 1-x; a:=0; b:=l; N:=[4, 10,20,50]; P := [seq( RiemannSum( f(x), x=a..b, partition=n, method=random, output=plot ), n=N )]: display ( P, insequence=true ); 89-92. Example CAS commands: Maple : with( Student[Calculusl] ); f := x -> sin(x); a:=0; b := Pi; plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); N:=[ 100, 200, 1000]; # (b) for n in N do Xlist := [ a+l.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: for n in N do # (c) Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); FunctionAverage(f(x),x=a..b,output=plot); # (d) fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); 83-92. Example CAS commands: Mathematica : (assigned function and values for a, b, and n may vary) Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands Clear[x, f, a, b, n] Copyright (c) 2006 Pearson Education Section 5 .4 The Fundamental Theorem of Calculus 315 {a, b}={0, tt}; n =10; dx = (b - a)/n; f=Sin[x] 2 ; xvals =Table[N[x], {x, a, b — dx, dx}]; yvals = f/.x — ► xvals; boxes = MapThread[Line[{{#l,0},{#l, #3}, {#2, #3), {#2, 0}]&,{xvals, xvals + dx, yvals}]; Plot[f, {x, a, b}, Epilog — ► boxes]; Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of commands. Clear[x, f, a, b, n] {a, b}={0, tt}; n =10; dx = (b - a)/n; f=Sin[x] 2 ; xvals =Table[N[x], {x, a + dx, b, dx}]; yvals = f /.x — > xvals; boxes = MapThread[Line[{{#l,0},{#l, #3}, {#2, #3), {#2, 0}]&,{xvals - dx,xvals, yvals}]; Plot[f, {x, a, b}, Epilog — » boxes]; Sum[yvals[[i]] dx, {i, l,Length[yvals]}]//N Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands. Clear[x, f, a, b, n] {a, b}={0, tt}; n =10; dx = (b - a)/n; f=Sin[x] 2 ; xvals =Table[N[x], {x, a + dx/2, b - dx/2, dx}]; yvals = f /.x — > xvals; boxes = MapThread[Line[{{#l,0},{#l, #3}, {#2, #3}, {#2, 0}]&,{xvals - dx/2, xvals + dx/2, yvals}]; Plot[f, {x, a, b},Epilog — > boxes]; Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N 5.4 THE FUNDAMENTAL THEOREM OF CALCULUS 133 4 1. J 2 (2x + 5) dx = [x 2 + 5x]! 2 = (0 2 + 5(0)) - ((-2) 2 + 5(-2)) = 6 2- £ (5 - 1) dx = [5x - f ] ^ = (5(4) - f ) - (5(-3) - ^) 3. r(3x-f)dx=[f-fl];=(^-il)-(^-f)=8 4. fjx 3 - 2x + 3) dx = [£ - x 2 + 3x] [ = (f - 2 2 + 3(2)) - (^ - (-2) 2 + 3(-2)) 12 ;i + D-o = i 5. X'( X 2 + V^)dx=[f + Ix 3 / 2 ]; 6. J o 5 x 3 / 2 dx = [2 x 5 / 2 ] I = | (5) 5 / 2 - = 2(5) 3 / 2 = 10^ 7. J" x-^ dx = [-5X-V5] » = (_§)_ (-5) = 1 8 - £'* dx = £' 2x " dx = t- 2x "] = 2 = (3) - (-2) = 1 Copy |t (c| 1 Pearson Etoatioo, Inc., publishing as Pearson Addison-Wesle 316 Chapter 5 Integration 9. J sinxdx= [-cosx]^ = (-costt)-(-cosO) = -(-1)- (-1) = 2 nir 10. 1(1+ cos x) dx = [x + sin x]J = (n + sin 7r) — (0 + sin 0) = 7r 11. J 2sec 2 xdx = [2tanx]g /3 = (2tan(f)) -(2tan0) = 2 v / 3-0 = 2 v / 3 12. J csc 2 xdx = [-cotx] 5 ^ = (-cot(f )) - (-cot(f)) = - (-V3) - (-73) = 2^3 2=0 13. J esc cot d0 = [-esc 0]^ = (-csc(^)) - (-esc (f)) = -^2- 14. J 4 sec u tan u du = [4 sec u]o /3 = 4 sec (f ) - 4 sec = 4(2) - 4(1) = 4 15 f ktsa* dt= f (1 + i cos2t) dt= [lt+ 1 sin2t]° /2 = (1(0)+ i sin 2(0)) - (1 (|) + \ sin 2(f)) 16. £l^f^ dt = £«(5 - I cos 2t ) dt = [I t - \ sin 2t] ^ ; ^(I-|cos2t)dt=[it-isin2t] ; (I (f ) - ^sm 2 (f )) - (I (_ f ) - 1 sin 2 (- f )) = - i sin ^ + I + i sin (=^ - = - ^ 6 4 "" 3 ' 6 ' 4 Ji " V 3 y 3 4 17. £^ /2 (8y 2 + sin y) dy = [ 7I-/2 8y 3 " M(f) 3 ,A /8(-l) 3 -r - cos y\ _ w/2 = (3 - cos I ) - I -V 1 - 18. 7T/4 ,< (4 sec 2 1 + 7rt~ 2 ) dt = [4 tan t - j] '^L 19. 20. X-7T/ -ir/3 = (4 tan (- I) - ^) - (4 tan ( f ) - ^) = (4(-l) + 4) - (4 (-^3) + 3) = 4^3 - 3 Jj (r+l) 2 dr=J i (r 2 + 2r+l)dr= [^+r 2 + r] _1 = (^£ + (-l) 2 + (-l)) - (| + l 2 +l) -VI / (t + 1) (t 2 + 4) dt = J_^(t 3 + t 2 + 4t + 4) dt = [I + f + 2t 2 + = (l^ + l^ + 2(y^ + v^ 16 w 2i -^-^) d -^- u i d -[^^]^(^w)-( .0* - & dv = /a 3 - v ") dv = [§a + 3^] u = (2W + w) - (# + 3#) 22. 23. //^ ds = J^(l + s-3/ 2 ) ds = [s - X] ^ = i ,/2 > 1 - -4- ) = \/2 - 2 3 / 4 + 1 ^ - Vs + 1 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 5 .4 The Fundamental Theorem of Calculus 317 XV 1/2 - 1) du = [2^ - u] I = (2^4 - 4) - (2^9 - 9) = 3 25j;jxid X =/;jxidx + /;ixidx=-/\dx + /;xdx = [-i]°_ 4+ [f]^(-f + ^) + (f-f) 24 r xA du 16 26. I | (cos x + |cos x| ) dx = I |(cos x + cos x) dx + I | (cos x — cos x) dx = I cos x dx = [sin x]g sin I — sin = 1 Jo .V* ^ => 4 f x 27. (a) I cos t dt = [sin t]j = sin ^/x - sin = sin y/x => ^ ( I cos t dt ) = ^ (sin a/x) = cos ^x (| x~ 1/2 ) v^ 2^1 (b) ;t (X ™ Sldl 7* (cos v^) (^ (v^)) = (cos ^x) (I x-V 2 ) = ^ 28. (a) J 3t 2 dt= [t 3 ]i° x = sin 3 x- 1 ^ ^ (X 3t 2 dt ] = ^ (sin 3 x - 1) = 3 sin 2 x cos x (b) i ( X 3t 2 dt J = (3 sin 2 x) (£ (sin x)) = 3 sin 2 x cos x 29. (a) f^du=f\^du=[j^Y i = j(t*f 2 -0=l? => |(/ V du I = |(|t 6 ) = 4t 5 (b) X v^ du) = 7? (| (t 4 )) = t 2 (4t 3 ) = 4t 5 30. (a) J sec 2 y dy = [tany]™ 9 = tan (tan 0) -0 = tan (tan 0) =$.±11 sec 2 y dy 1= ± (tan (tan 0)) (sec 2 (tan 0)) sec 2 / n lan \ (b) ± (J sec 2 y dy = (sec 2 (tan 9)) (± (tan 9)) = (sec 2 (tan 0)) sec 2 31. y = / Vi+T 2 dt=^ £ = v/r 32 -y = fi dt => ^ = I , x > dx x ' 33. y = I sint 2 dt= - / A sin t 2 dt 2 Ht -^ ^y (^(v^) 2 ) (£ (v^)) = -(sinx)(|x-V2) = _^ 34. y = X cos dt => I = (cos a/x 2 ) (5 (x 2 )) = 2x cos |x 35. y /. M < f =-> £ = l . 1 .„ (£(sinx)) = -* (cosx) x/T^ ' ' ' 2 dx ~ x/l-sin^x Vdx ««x = co^x = j since II < 1 COS X COS X 112 Jo 36. y = .l. TT? =* Mli;)(l("")) = (iH)(»'') = ' Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 318 Chapter 5 Integration 37. -x 2 - 2x = => -x(x + 2) = =4> x = or x = -2; Area = -J (-x 2 - 2x)dx + J_ 9 (-x 2 - 2x)dx - J (-x 2 - 2x)dx ((- _ (- 2 ) 3 3 °?-0 2 3 u (-2) 2 )-(-^-(-3) 2 )) (-2) 3 (-2)' -^-2 2 W-f-0 2 ))=f / H3 A2 \ / -l -2 1 y = -x 2 - 2x -4 -6 -8 \ J \ i 38. 3x 2 — 3 = =^ x 2 = l =>- x= ±1; because of symmetry about they-axis, Area = 2 ( J (3x 2 - 3)dx + J (3x 2 - 3)dx 2 (- [x 3 - 3x] I + [x 3 - 3x] I) = 2 [- ((l 3 - 3(1)) - (0 3 - 3(0))) + ((2 3 - 3(2)) - (l 3 - 3(1))] = 2(6) = 12 y = 3x -3 39. x 3 - 3x 2 + 2x = => x (x 2 - 3x + 2) = => x(x - 2)(x - D = => x = 0, 1, or 2; J (x 3 - 3x 2 + 2x)dx - J (x 3 - 3x 2 + 2x)dx Area \ -x 3 + x 2 - K -x 3 + x 2 (^-i 3 + i 2 )-(^-o 3 + o 2 ) (^-2 3 +2 2 )-(^-l 3 +l 2 )]=I y - x - 3x + 2x 40. x 3 - 4x = => x (x 2 - 4) = =>■ x(x - 2)(x + 2) = /0 n2 (x 3 - 4x)dx - I (x 3 - 4x)dx 2x< 2x< o 1 2(0) 2 ) (tfi - 2(-2) 2 ) - [{* - 2(2) 2 ) - (* - 2(0) 2 )] 41. x 1 / 3 =0 =^ x = 0;Area= - J x 1 / 3 dx + J x 1 / 3 dx = [_3 x 4/3]0 + [3 x 4/3j 8 = (~ I (0) 4/3 )- (" 1 (-1°) 4/3 ) + Q (8) 4 / 3 ) - (1 (O) 4 / 3 ) 51 4 Copyright (c) 2006 Pearson Education, Inc Section 5.4 The Fundamental Theorem of Calculus 319 42. x 1 / 3 - x = => x 1 / 3 (l-x 2 / 3 ) -- = => xV 3 = Oor 1 - x 2 / 3 = => x = = or 1 = x 2 / 3 => ; t = or 1 = x 2 => x = or ± i; Area = - 1 (x 1 / 3 - x ) dx +r ( x l/3 . - x) dx - r = -[l- 4/3 -f]° i + [|x 4 / 3 - 2 Jo -[§x« _ x^ = - [(l(0) 4 / 3 -f )-(K-D 4/3 - k ■n + [(!d) 4/3 -f) -(|(0) 4 / 3 - ?)] -[(!(8) 4/3 -f) -(i(D 4/3 - «)] = I + |-(-20- 3 , 1\ 83 4 ' 2) ~ 4 .1/3 >c)dx y 1/3 y = x -x \2 4 6 8 -l | -2 | -3 -4 >. | -5 >v 1 -6 N 43. The area of the rectangle bounded by the lines y = 2, y = 0, x = ir, and x = is 27r. The area under the curve y = 1 + cos x on [o^iis/; (1 + cos x) dx = [x + sin x][J = (ir + sin ir) — (0 + sin 0) = it. Therefore the area of the shaded region is 2tt — it = ir. 44. The area of the rectangle bounded by the lines x=|,x=^,y = sin| = i = sin^, and y = is 1 (5n _ 7T\ _ 7T 2 V 6 6) 3 n 57r/6 ir/6 . The area under the curve y = sin x on [|, ^] is I sin x dx = [—cos x]J, (—cos ^) — (—cos |) = — (— 2i^~2 = V 3' Therefore the area of the shaded region is \J 3 — | . 45. On [- f ,0] : The area of the rectangle bounded by the lines y = \fl, y = 0, 9 = 0, and 9 = - f is \fl (f ) -7T/4 = ^(— . The area between the curve y = sec 6 tan 6 and y = is — I sec 8 tan A9 = [—sec 9]° = (—sec 0) — (—sec (— f )) = v2 — 1. Therefore the area of the shaded region on [— |,0] is ^ — h ( v 2 — 1 j On [0, f ] : The area of the rectangle bounded by 8 = J , 9 = 0, y = \fl, andy = is \fl (f ) = ^ . The area under the curve y = sec 9 tan 9 is I sec 8 tan 9 dd = [sec #]q = sec | — sec = y 2 — 1 . Therefore the area of the shaded region on [0, j\ is ^j (^ + V / 2-l) + (^- V / 2+l) 2 — 1 ) . Thus, the area of the total shaded region is 7IV2 46. The area of the rectangle bounded by the lines y = 2, y = 0, t = — ^, and t = 1 is 2 (l — (— J) J = 2+ f . The area under the curve y = sec 2 1 on [— |, 0] is I sec 2 1 dt = [tan t]^_, 4 = tan — tan (— |) = 1. The area under the curve y = 1 - t 2 on [0, 1] is J ( 1 - t 2 ) dt = 1 1 - |1 = M - y J area under the curves on [—J, l] is 1 Therefore the area of the shaded region is (2 - V ) - ( - f ) = I . Thus, the total 5 _ 1 , 7T 3 ~~ 3 t 2 47. y = J i dt - 3 =4> ^ = \ and y(?r) = J \ dt - 3 = - 3 = -3 =4> (d) is a solution to this problem. 48. y = J sectdt + 4 => ^|= sec x and y(-l) = J sec t dt + 4 = + 4 = 4 =4> (c) is a solution to this problem. 49. y = I sec t dt + 4 => ^ = sec x and y(0) = I sec t dt + 4 = + 4 = 4 => (b) is a solution to this problem. Cop # (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 320 Chapter 5 Integration 50. y = J 1 dt - 3 => ^ = I and y(l) = J i dt - 3 = - 3 = -3 => (a) is a solution to this problem 51. y = J sectdt + 3 52. y= J v 7 ! + t 2 dt - 2 53. s = / f(x) dx + s 54. v = I g(x) dx + v " to *> <0 55. Area=J^(h-(g)x 2 )dx=[hx """^ 4hx 3 3b 2 b\ 4h(|)- 2 y 3b 2 b\ 4h(-|)' 2/ 3b 2 -bh w^ (_bh + bhx bh _bh = 2 bh 2 6 3 — 3 h y = h - (4h/b 2 )x* -b/2 56 - r =X( 2 -^ ) dx = 2 r( i -(^w) dx = 2 [ x -(^T)]o= 2 [( 3+ (^))-(° + ^)] 2 [3 \ - 1] = 2 (2 i) = 4.5 or $4500 57- | = 5$5 = I ^ 1/2 =* ^ = Jj t^^dt = [tV] ; = ^ c(100) - c(l) = yiOO -\[\ = $9.00 58. By Exercise 57, c(400) - c(100) '400 100 = 20- 10 = $10.00 59. (a) v = | = £ J f(x) dx = f(t) =>• v(5) = f(5) = 2 m/sec (b) a = ^j is negative since the slope of the tangent line at t = 5 is negative (c) s = I f(x) dx = | (3)(3) = | m since the integral is the area of the triangle formed by y = f(x), the x-axis, and x = 3 (d) t = 6 since from t = 6 to t = 9, the region lies below the x-axis (e) At t = 4 and t = 7, since there are horizontal tangents there (f) Toward the origin between t = 6 and t = 9 since the velocity is negative on this interval. Away from the origin between t = and t = 6 since the velocity is positive there. (g) Right or positive side, because the integral of f from to 9 is positive, there being more area above the x-axis than below it. 60. (a) v = f = | J q g(x) dx = g(t) => v(3) = g(3) = m/sec. (b) a = ^ is positive, since the slope of the tangent line at t = 3 is positive (c) At t = 3, the particle's position is J g(x) dx = | (3)(-6) = -9 f 6 (d) The particle passes through the origin at t = 6 because s(6) = | g(x) dx = (e) At t =7, since there is a horizontal tangent there (f) The particle starts at the origin and moves away to the left for < t < 3. It moves back toward the origin for 3 < t < 6, passes through the origin at t = 6, and moves away to the right for t > 6. (g) Right side, since its position at t = 9 is positive, there being more area above the x-axis than below it at t = Cop # (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 5.4 The Fundamental Theorem of Calculus 321 61. k > => one arch of y = sin kx will occur over the interval [0, |] =>- the area = I sin kx dx = [— j cos kx] 7r/k -E cos ( k (!))-(-E cos (0)) 62 - SS?/o"A dt = lim /o^ dt lim-M**- = lim^T-A x^O x^0° oc. 63. 64. J f(t) dt = x 2 - 2x + 1 => f(x) = £ J f(t) dt = £ (x 2 - 2x + 1) = 2x - 2 I f(t) dt = x cos 7rx =>■ f(x) = gj I f(t) dt = cos 7rx — 7rx sin 7rx => f(4) = cos 7r(4) — 7r(4) sin 7T(4) = 1 J'X+1 /•!+! ^dt => f' W = _ TT ^ = -zL =, f'(l) = -3;f(l) = 2-J ^=2-0 = 2; I 2 i "I- i 1 + (x + 1) L(x) = -3(x - 1) + f(l) = -3(x - 1) + 2 = -3x + 5 r*x 2 66. g(x) = 3 + J sec(t- 1) dt => g'(x) = (sec (x 2 - l))(2x) = 2x sec (x 2 - 1) => g'(-l) = 2(-l) sec((-l) 2 - 1) r (-i) 2 pi = -2;g(-l)= 3 + J sec(t- 1) dt = 3 + J sec(t- 1) dt = 3 + = 3; L(x) = -2(x- (-l)) + g(-l) = -2(x+ l) + 3 = -2x+ 1 67. (a) True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus. (b) True: g is continuous because it is differentiable. (c) True, since g'(l) = f(l) = 0. (d) False, since g"(l) = f'(l) > 0. (e) True, since g'(l) = and g"(l) = f'(l) > 0. (f) False: g"(x) = f'(x) > 0, so g" never changes sign. (g) True, since g'(l) = f(l) = and g'(x) = f(x) is an increasing function of x (because f'(x) > 0). 68. (a) True: by Part 1 of the Fundamental Theorem of Calculus, h'(x) = f(x). Since f is differentiable for all x, h has a second derivative for all x. (b) True: they are continuous because they are differentiable. (c) True, since h'(l) = f(l) = 0. (d) True, since h'(l) = and h"(l) = f'(l) < 0. (e) False, since h"(l) = f'(l) < 0. (f) False, since h"(x) = f'(x) < never changes sign. (g) True, since h'(l) = f(l) = and h'(x) = f(x) is a decreasing function of x (because f'(x) < 0). 69. = cosx 70. The limit is 3x 2 -h-0.5 Cop # (c) 1 Pearson Ediratio^, k, publishing as Pearson Addison-Wesle 322 Chapter 5 Integration 71-74. Example CAS commands: Maple : with( plots ); f := x -> x A 3-4*x A 2+3*x; a:=0; b:=4; F := unapply( int(f(t),t=a..x), x ); # (a) pi := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#71(a) (Section 5.4)" ): pi; dF := D(F); # (b) ql := solve( dF(x)=0, x ); ptsl := [ seq( [x,f(x)], x=remove(has,evalf([ql]),I) ) ]; p2 := plot( ptsl, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where F '( x )=0" ): display( [pl,p2], title="71(b) (Section 5.4)" ); incr := solve( dF(x)>0, x ); # (c) deer := solve( dF(x)<0, x ); df := D(f); # (d) p3 := plot( [df(x),F(x)], x=a..b, legend=["y = f '(x)","y = F(x)"], title="#71(d) (Section 5.4)" ): p3; q2 := solve( df(x)=0, x ); pts2 := [ seq( [x,F(x)], x=remove(has,evalf([q2]),I) ) ]; p4 := plot( pts2, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where f '00=0" ): display( [p3,p4], title="71(d) (Section 5.4)" ); 75-78. Example CAS commands: Maple : a:= 1; u := x -> x A 2; f :=x->sqrt(l-x A 2); F := unapply( int( f(t), t=a..u(x) ), x ); dF := D(F); # (b) cp := solve( dF(x)=0, x ); solve( dF(x)>0, x ); solve( dF(x)<0, x ); d2F := D(dF); # (c) solve( d2F(x)=0, x ); plot( F(x), x=-l..l, title="#75(d) (Section 5.4)" ); 79. Example CAS commands: Maple : f:=T; ql := Diff( Int( f(t), t=a..u(x) ), x ); dl := value( ql ); 80. Example CAS commands: Maple : f:=T; q2 := Diff( Int( f(t), t=a..u(x) ), x,x ); Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 5.5 Indefinite Integrals and the Substitution Rule 323 value( q2 ); 71-80. Example CAS commands: Mathematica l (assigned function and values for a, and b may vary) For transcendental functions the FindRoot is needed instead of the Solve command. The Map command executes FindRoot over a set of initial guesses Initial guesses will vary as the functions vary. Clear[x, f, F] {a, b}= {0, 2tt}; f[x_] = Sin[2x] Cos[x/3] F[x_] = Integrate [f[t], {t, a, x}] Plot[{f[x],F[x]),{x, a, b}] x/.Map[FindRoot[F'[x]==0, {x, #}] &,{2, 3, 5, 6}] x/.Map[FindRoot[f [x]==0, {x, #}] &,{ 1, 2, 4, 5, 6}] Slightly alter above commands for 75 - 80. Clear[x, f, F, u] a=0; f[x_] = x 2 - 2x - 3 u[x_] = 1 — X 2 F[x_] = Integrate [f[t], {t, a, u(x)}] x/.Map[FindRoot[F'[x]==0,{x,#}]&,{l,2,3,4)] x/.Map[FindRoot[F"[x]==0,{x,#}] &,{ 1, 2, 3, 4}] After determining an appropriate value for b, the following can be entered b = 4; Plot[{F[x],{x,a,b}] 5.5 INDEFINITE INTEGRALS AND THE SUBSTITUTION RULE 1 . Let u = 3x => du = 3 dx => | du = dx I sin 3x dx = I | sin u du = — | cos u + C = — | cos 3x + C 2. Let u = 2x 2 =>- du = 4x dx =>■ \ du = x dx I x sin (2x 2 ) dx = / I sin u du = — | cos u + C = — j cos 2x^ + C 3. Let u = 2t =^ du = 2 dt => \ du = dt I sec 2t tan 2t dt = I | sec u tan u du = | sec u 4- C = \ sec 2t + C 4. Let u = 1 — cos ^ =>- du = \ sin | dt => 2 du = sin | dt J(l -cos |) 2 (sin i) dt= J 2u 2 du = § u 3 + C = § (l -cos |) 3 + C 5. Let u = 7x — 2 =4> du = 7 dx => ^ du = dx J" 28(7x - 2y 5 dx = J" i (28)u~ 5 du = J" 4u~ 5 du = -u~ 4 + C = -(7x - 2)~ 4 + C 6. Let u = x 4 - 1 => du = 4x 3 dx =^ \ du = x 3 dx / x 3 (x 4 - l) 2 dx = J \ u 2 du = si + C = i (x 4 - l) 3 + C Copyright (c) 2006 Pearson Education 324 Chapter 5 Integration 7. Let u = 1 - r 3 =>• du = -3r 2 dr => -3 du = 9r 2 dr f 9fd^ = r _ 3u -i/2 du = _ 3 (2) u i/2 + C =-6(1- r 3 ) 1/2 + C 8. Let u = y 4 + 4y 2 + 1 => du = (4y 3 + 8y) dy => 3 du = 12 (y 3 + 2y) dy J" 12 (y 4 + 4y 2 + l) 2 (y 3 + 2y) dy = J 3u 2 du = u 3 + C = (y 4 + 4y 2 + l) 3 + C 9. Let u = x 3 / 2 - 1 =>■ du = § x 1 / 2 dx =>■ | du = y/x dx J ,/x sin 2 (x 3 / 2 - 1) dx = J | sin 2 u du = | (§ - i sin 2u) + C = i (x 3 / 2 - l) - 1 sin (2x 3 / 2 - 2) + C 10. Let u = -i =4> du = - 2 dx I ^ cos 2 (i) dx = / cos 2 (— u) du = I cos 2 (u) du J + isin2u)+C = -^ + isin(-; £-*•*(!) +c C = - 1 cot 2 26> + C 11. (a) Let u = cot 261 =>■ du = -2 esc 2 20 d0 => - 1 du = esc 2 20 d0 J esc 2 20 cot 20 d0 = - / \ u du = - \ (| ) + C = - f (b) Let u = esc 20 =>• du = -2 esc 20 cot 20 d0 => - \ du = esc 20 cot 20 d0 J esc 2 20 cot 20 d0 = J - 1 u du = - \ (| ) + C = - f + C = - \ esc 2 20 + C 12. (a) Let u = 5x + 8 => du = 5 dx =>■ | du = dx /7fe5 = /K7u) du =i/ u " /2du =H2u 1/2 )+C=lu 4 / 2 + C=| v ^T8 + C (b) Let u = v/5x + 8 =4> du = ± (5x + 8)" 1 / 2 (5) dx ^ § du /5x+8 13. Let u = 3 - 2s => du = -2 ds =4> - \ du = ds / ^3^ ds = / ^u" (- \ du) = - \ JV/ 2 du = (- \) (| u 3 / 2 ) + C = - \ (3 - 2s) 3 / 2 + C 14. Let u = 2x + 1 => du = 2 dx =$■ \ du = dx / (2x + l) 3 dx = / u 3 (i du) = \ J u 3 du = (I) (£) + C = | (2x + l) 4 + C 15. Let u = 5s + 4 => du = 5 ds =>• ± du = ds /^ds = /^(Idu)=i/u-V 2 du=(i)(2uV 2 ) +C =|y^T^ + C 16. Let u = 2 — x =>- du= — dx => — du = dx J(2^dx = /^ = -3/u- 2 du=-3(^) + C=^ + C 17. Let u = 1 - 2 =4> du = -20 d0 => - 1 du = d0 / VT 3 ^ 2 d0 = / \/u" (- 1 du) = - I / u 1 / 4 du = (- I) (I u 5 / 4 ) + C = - § (1 - 2 ) 5/4 + C 18. Let u = 2 - 1 => du = 20 d0 => 4 du = 80 d0 / 80 Vfl 2 - 1 66 = f Vu"(4 du) = 4 / u 1 / 3 du = 4 (| u 4 / 3 ) + C = 3 (0 2 - 1) 4/3 + C Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 5.5 Indefinite Integrals and the Substitution Rule 325 19. Let u = 7 - 3y 2 => du = -6y dy => - \ du = 3y dy /3 y y7^3?dy = /^(-idu) =-i/u 1 /' 2 du= (-i) (f u 3 / 2 ) + C = - \ (7 - 3y 2 ) 3/2 + C 20. Let u = 2y 2 + 1 => du = 4y dy / 4y dy V^y 2 + 1 J 4- du = J u- 1 / 2 du = 2U 1 / 2 + C = 2v/2y 2 + 1 + C 21. Let u = 1 + Jx =>■ du = 4r dx => 2 du = 4" v ^v x v x dx r i d X = r ^ 22. Let u = 1 i+\A + C du J^^dx = /u 3 (2du) = 2(iu 4 )+C=i(l -V dx => 2 du = -4- dx 23. Letu = 3z + 4 =>• du = 3 dz => 5 du = dz J cos (3z + 4) dz = I (cos u) (I du) = 5 J cos u du = | sin u + C = | sin (3z + 4) + C 24. Let u = 8z - 5 =4> du = 8 dz => | du = dz I sin (8z — 5) dz = I (sin u) ( I du) = | I sin u du = | (—cos u) + C = — | cos (8z — 5) + C 25. Let u = 3x + 2 =4> du = 3 dx => ± du = dx J sec 2 (3x + 2) dx = J (sec 2 u) (5 du) = \ J sec 2 u du = | tan u + C = | tan (3x + 2) + C 26. Let u = tan x =>■ du = sec 2 x dx J tan 2 x sec 2 x dx = J u 2 du = | u 3 + C = 5 tan 3 x + C 27. Letu = sin (|) =4> du = | cos (|) dx =>- 3 du = cos (|) dx f sin 5 (f ) cos (I) dx = / u 5 (3 du) = 3 (± u°) + C = \ sin G (f ) + C 28. Let u = tan (|) => du = \ sec 2 (|) dx =^ 2 du = sec 2 (|) dx /tan 7 (I) sec 2 (I) dx = J u 7 (2 du) = 2 (± u 8 ) +C= 5 tan 8 (f) +C 29. Let u = f| - 1 =^ du = | dr =>• 6 du = r 2 dr J r 2( f i„ 1 ) 5 dr= J u 5 (6du) = 6 J u 5 du = 6 ^) +C= ^_ 1 ) 6 +C 30. Let u = 7 - ^ => du = - \ r 4 dr => -2 du = r 4 dr J r 4( 7 _ { i) 3 dr= J u 3 ( _2du) = -2/u 3 du=-2(^)+C=-l(7-ii) 4 + C 31. Let u = x 3 / 2 + 1 => du = | x 1 / 2 dx =>■ | du = x 1 / 2 dx J x 1 / 2 sin (x 3 / 2 + 1) dx = J (sin u) (| du) = § J sin u du = § (-cos u) + C = - \ cos (x 3 / 2 + l) + C Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 326 Chapter 5 Integration du = | x 1 / 3 dx 32. Let u = x 4 / 3 - 8 J x 1 / 3 sin (x 4 / 3 - 8) dx = J (sin u) | du = x 1 / 3 dx du / sin u du =| (—cos u) + C cos X 4/3 33. Let u = sec (v + |) => du = sec (v + f ) tan (v + f ) dv J sec (v + |) tan (v + |) dv = J du = u + C = sec (v + |) + C 34. Letu = csc(^) => du = -icsc(^) cot(^) dv => -2du = csc(^) cot(^) dv / esc (^j cot (^ *■) dv = J -2 du = -2u + C = -2 esc i ^)+C 35. Let u = cos (2t +1) =^ du = -2 sin (2t + 1) dt => - | du = sin (2t + 1) dt f ^ffttn * = f - i ^ = j- + C J cos 2 (2t+l) J 2 u 2 2u 1 2cos(2t+ 1) c 36. Let u = 2 + sin t => du = cos t dt / 6 cos t (2 + sin t) = dt : / | du = 6 / u~ 3 du = 6 (^|) + C = -3(2 + sin t)~ 2 + C 37. Letu = coty =>- du = — esc 2 y dy =>- — du = esc 2 y dy J v/coTy esc 2 y dy = J v /u(-du) = - j u 1 / 2 du = - § u 3 / 2 + C = - f (cot y) 3 / 2 + C |(cot 3 y) 1/2 + C 38. Let u = sec z => du = sec z tan z dz / dz J i du = / u- 1 / 2 du = 2u J / 2 + C = 2^^^ + C 39. Letu 1 = t _1 - 1 =^ du = -t~ 2 dt => -du = i dt / p cos (i — l) dt = I (cos u)(— du) = — I cos u du = —sin u + C = —sin Q — l) + C 40. Let u = + 3 = t 1 / 2 + 3 => du = 1 1" 1 / 2 dt => 2 du = X dt J 4j cos (i/t + 3) dt = J (cos u)(2 du) = 2 J cos u du = 2 sin u + C = 2 sin (i/t + 3) + C 41. Let u = sin \ => du = (cos j) (- gj) d# =4> -du = p cos | d# J p sin i cos | d6» = J -u du = - | u 2 + C = - i sin 2 | + C 42. Let u = esc \f~6 => du = [-esc \/~6 cot y/Fj (~Krj d6* =>• -2 du = -4- cot \f~6 esc y^ d6> f /T wtt d <? = f 4s cot y/d esc y/9d6= f -2 du = -2u + C = -2 esc v^ + C = J Vf? sin 2 \/0 J vtf J sin 43. Let u = s 3 + 2s 2 - 5s + 5 => du = (3s 2 + 4s - 5) ds JV + 2s 2 - 5s + 5) (3s 2 + 4s - 5) ds = / u du = f + C = ( s3 + 2s2 - 5s + 5 > 2 + C 44. Let u = d i - 26 2 + W - 2 => du = (46> 3 - 46 + 8) dd =>• 1 du = (6» 3 - 6> + 2) d<9 / (0 4 - 20 2 + 89 - 2) (0 3 - 6 + 2) d# = / u (| du) = \ J u du = \ (f ) + C = ( g4 - 2fl2 + 8g - 2 ) 2 + C Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 5.5 Indefinite Integrals and the Substitution Rule 327 45. Let u = 1 + t 4 =>• du = 4t 3 dt =^ \ du = t 3 dt Jt 3 (l+t 4 ) 3 dt=/u 3 (|du) HX) C=i(l+t 4 ) 4 + C 46. Let u = 1 - 1 => du = 4j dx ^dx J# dx =J? /^JTTidx = /^du = /uV 2d u=|u3/ 2 + C=|(l-I) 3/2 + C 47. Let u = x 2 + 1. Then du = 2xdx and |du = xdx and x 2 = u — 1. Thus I x 3 y x 2 + 1 dx = / (u — l)|-^/u = i/(u 3 / 2 - u 1 / 2 )du = i [|u 5 / 2 - |u 3 / 2 ] + C = |u 5 / 2 - |u 3 / 2 + C = |(x 2 + 1) 5/2 - i(x 2 + if' 2 du + C 48. Letu = x 3 + 1 =^ du = 3x 2 dx and x 3 = u - 1. So f3x 5 y/x. 3 + ldx= J (u - l)y / udu = J(u 3/2 - u 1 / 2 ) du 2 u 5/2 _ 2 u 3/2 + C = 2 (x 3 + ^5/2 _ 2,3 + ^3/2 + c (a) Let u = tan x => du = sec 2 x dx; v = u 3 =>• dv = 3u 2 du =>• 6 dv = 18u 2 du; w = 2 + v =>• dw = dv / dx = f -^- 2 du = f ^^v = f 6*v = 6 f -2 d _ 6 -1 + C = _ ^_ + C J (2 + u 3 ) J (2 + v) 2 J w 2 J 2 + v 18 tan 2 x sec 2 x (2 + tan 3 x) 2 " A ~~ J (2 + u 3 ) 2 "" ~ J (2 + v) 2 = - 2+TP +C= — 2 + tan 3 x + ( ~' i3 v _v ^n 'J ton2 v e ~~2 v -J v — v. /: J.. lO *-or»2 „ „~~2 . (b) Let u = tan 3 x =>- du = 3 tan x sec x dx =>• 6 du = 18 tan x sec - x dx; v = 2 + u =>- dv = du J 1 8 tan 2 x sec 2 x j v / 6 du l 6 dv 6 i /-i 6 i /-i QX — J (2 + u) 2 _ J v 2 ■ - + L ' "" +L (2 + tan 3 x) 2 J (2 + u) 2 J v 2 v ' *- 2 + u ' *- 2 + tan 3 x (c) Let u = 2 + tan 3 x =>■ du = 3 tan 2 x sec 2 x dx =>• 6 du = 18 tan 2 x sec 2 x dx + C / 1 8 tan 2 x sec 2 x j v / 6 du _ 6 ■ p (2 + tan 3 x) 2 QX ~ J u2 - u + ^ 2 + tan 3 x 50. (a) Let u = x — 1 => du = dx; v = sin u => dv = cos u du; w = 1 + v 2 =>• dw = 2v dv =>• | dw = v dv I i/l + sin 2 (x — 1) sin (x — 1) cos (x — 1) dx = I y 1 + sin 2 u sin u cos u du = J vy 1 + v 2 dv = J" 1 yV dw = 1 w 3 / 2 + C = i (1 + v 2 ) 3/2 + C = i (1 + sin 2 u) 3/2 + C = i (1 + sin 2 (x - 1)) 3/2 + C (b) Let u = sin (x — 1) =>• du = cos (x — 1) dx; v = 1 + u 2 => dv = 2u du =>• | dv = u du J a/1 +sin 2 (x - 1) sin (x - 1) cos (x - 1) dx = J u yTTu 2 du=Jiy / vdv = Ji v 1 / 2 dv = (I (I) v3/2 ) + C = 5 v3/2 + C = 5 (1 + u 2 ) 3/2 + C = i (1 + sin 2 (x - 1)) 3/2 + C (c) Let u = 1 + sin 2 (x — 1) =>• du = 2 sin (x — 1) cos (x — 1) dx =>• \ du = sin (x — 1) cos (x — 1) dx f y/l + sin 2 (x- l)sin(x- l)cos(x- 1) dx = J±y / u~du = J \ u 1 / 2 du = \ (|u 3 / 2 ) +C = i(l + sin 2 (x- l)) 3/2 + C 51. Let u = 3(2r - l) 2 + 6 =>• du = 6(2r - 1)(2) dr => ^ du = (2r - 1) dr; v = y 7 ! 7 ^> dv = ^ du => ± dv du / l (2r- l)cos V3(2r-l) 2 + 6 dr / co^/uX / i v / 3(2r-l) 2 + 6 "* J V V 7 " ^ ' '" i sin v/3(2r- l) 2 + 6 + C du J (cos v) (i dv) = 1 sin v + C = \ sin y 7 ^ + C 52. Let u = cos y/~d => du = (-sin y^) f ^O d6» =>■ -2 du = 5^ d# f , sin ^ M= \ si "^ d# = f ^?# = -2 f u- 3 / 2 du = -2 (-2u- 1 /2) + C = 4 + C Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 328 Chapter 5 Integration y cos v 8 + c 53. Let u = 3t 2 - 1 => du = 6t dt =4> 2 du = 12t dt s = Jl2t(3t 2 - l) 3 dt= Ju 3 (2du) = 2(|u 4 ) +C= |u 4 + C= \ (3t 2 - 1) 4 + C; s = 3 when t=l => 3 = ± (3 - l) 4 + C =5> 3 = 8 + C =4> C = -5 => s = \ (3t 2 - l) 4 - 5 54. Let u = x 2 du = 2x dx =>• 2 du = 4x dx y = J 4x (x 2 + 8)~ 1/3 dx = J u- J /3 (2 du) = 2(1 u 2 / 3 ) + C = 3u 2 / 3 + C = 3 (x 2 y = when x = => = 3(8) 2 / 3 +C => C = -12 =^> y = 3(x 2 \ 2 /3 12 \ 2 / 3 C; 55. Letu = t I> du = dt / 8 sin 2 u du = 8 (jj - 1 sin 2u) + C = 4 (t + f0 - 2 sin (2t + § ) + C; 8 when t = 2 sin I => s = 4(t+f^) -2sin(2t + C =4> C = 8 4t-2sin(2t+ |) +9 56. Let u = | - 9 => -du = d6> r = J 3 cos 2 (| - 0) &6 = - J" 3 cos 2 u du = -3 (| + \ sin 2u) + C cos when = 3jr 8 c -§(f-0)-f sin (f-20)+C; r =-i(f-^) 3 3 29) + § | cos 26> + f + -20) + 57. Let u = 2t - S => du = 2 dt =>• -2 du = -4 dt 2 cos (2t - f ) | = J -4 sin (2t - |) dt = J (sin u)(-2 du) = 2 cos u + Ci = 2 cos (2t - §) + Ci at t = and g = 100 we have 100 = 2 cos (- §) + Ci => Ci = 100 => s =4> s = J* (2 cos (2t - |) + 100) dt = J (cos u + 50) du = sin u + 50u + C 2 = sin (2t - § at t = and s = we have = sin (- §) + 50 (- f ) + C 2 => C 2 = 1 + 25?r =>• s = sin (2t - f ) + 100t - 25tt + (1 + 25tt) => s = sin (2t - § ) + lOOt + 1 100 50(2t-|)+C 2 ; 58. Let u = tan 2x => du = 2 sec 2 2x dx =>- 2 du = 4 sec 2 2x dx; v = 2x => dv = 2 dx => h dv = dx !| = J* 4 sec 2 2x tan 2x dx = J" u(2 du) = u 2 + d = tan 2 2x + C x ; dy dx at x = and ^ = 4 we have 4 = + Ci => Ci = 4 => f£ = tan 2 2x + 4 = (sec 2 2x - 1) + 4 = sec 2 2x + 3 ^ y = J (sec 2 2x + 3) dx = J (sec 2 v + 3) (± dv) = \ tan v + § at x = and y = - 1 we have - 1 = ± (0) + + C 2 =>■ C 2 = - 1 =>• y C 2 = 5 tan2x+3x + C 2 ; i tan 2x + 3x - 1 59. Let u = 2t =» du = 2 dt =4> 3 du = 6 dt s = J 6 sin 2t dt = J (sin u)(3 du) = -3 cos u + C = -3 cos 2t + C; at t = and s = we have = -3 cos + C => C = 3 =4> s = 3 - 3 cos 2t s (|) =3 — 3 cos(7r) = 6 m 60. Let u = 7rt => du = 7r dt => tt du = 7r 2 dt v = I 7r 2 cos 7rt dt = I (cos u)(7r du) = tt sin u + Ci = 7r sin (7rt) + Ci ; at t = and v = 8 we have 8 = 7r(0) + Ci =4> Ci = 8 => v = ^ = n sin (7rt) = J sin u du + 8t + C 2 = -cos (7rt) + 8t + C 2 ; at t = and s = we have = ! => s = J (tt sin (7rt) + 8) dt •1+Ca =>■ C 2 = 1 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 5.6 Substitution and Area Between Curves 329 S = 8t - COS (7Tt) +1 => s(l) = 8 - COS 7T + 1 = 10 m 61. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover, sin 2 x + Ci = 1 - cos 2 x + Ci => C 2 = 1 + Ci ; also -cos 2 x + C 2 ,2.x c 2 Ci 62. Both integrations are correct. In each case, the derivative of the function on the right is the integrand on the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover, tajp + C sec x— 1 /-i sec x a constant 63. (a) (xTq) J V max sin 12U7rt dt = 60 [-V„ (j^) cos (1207rt)] J 760 = - ¥=■ [cos 2tt - cos 0] = -^[l-l]=0 (b) V max = \[2N m = ^(240) w 339 volts (c) J (V max ) 2 sin 2 1207rtdt = (V max ) 2 J 1/60 _ (V,,,,,) 2 1 - cos 240-tt t ■> dt (v- - ,.1/60 fL J g ( (1 - cos 2407rt) dt t Ilk) sin240^t Jo [(§13 ~ (21k) sin ( 4 ^)) - (0- (21k) s in (0))] = ^ 20 5.6 SUBSTITUTION AND AREA BETWEEN CURVES 1. (a) Letu = y+1 => du = dy; y = => u = 1, y = 3 =4> u = 4 J 3 VV+i d y = T ul/2 du = [f u3/2 ] 1 = (I) (4)3/2 ~ (I) (1)3/2 (8) :u<i)=£ (b) Use the same substitution for u as in part (a); y = — 1 =>• u = 0, y = => u = 1 £y>7TTdy = X 1 uV 2 du=[lu3/ 2 ];=(|)(l)3/ 2 -0 2. (a) Let u = 1 - r 2 => du = -2r dr => - 5 du = r dr; r = =>• u=l,r=l =>• u = £rv / T^ 2 dr= ^°-i v A 1 du= [- | u 3 / 2 ] J = - (- ±) (l) 3 / 2 = § (b) Use the same substitution for u as in part (a); r = — 1 =>• u = 0, r = 1 => u = J_ 1 i r/r^dr=X°-i v AIdu = 3. (a) Let u = tan x =>• du = sec 2 x dx; x = => u = 0, x = f =>• u=l /W 4 I t Jo tan x sec - x dx Jo du S-o (b) Use the same substitution as in part (a); x ■ l,x = =>• u = I" <J — 7rA tan x sec x dx /> du 4. (a) Let u = cos x => du = - I 3 cos 2 x sin x dx = I sin x dx => — du = sin x dx; x = => u = 1, x = 7r => u -1 -3u 2 du = f-u 3 l 7 1 (b) Use the same substitution as in part (a); x = 2tt -3u 2 du = 2 _(_1)8 _ (_ ( 1)8) = 2 U=1,X=37T =>■ U = — 1 J 3 cos 2 x sin x dx = I 2-k J 1 Copffigl (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle 330 Chapter 5 Integration 5. (a) u = 1 + t 4 => du = 4t 3 dt => \ du = t 3 dt; t = => u = 1, t = 1 =>• u = 2 21 _ il _ I! 16 16 16 (b) Use the same substitution as in part (a); t = — 1 =>■ u = 2, t = 1 =>• u = 2 J t 3 (1 + t 4 ) 3 dt = j \ u 3 du = =>■ du = 2t dt =^> i du = t dt; t = => u = 1, t = \fl •J u «/ 1 (b) Use the same substitution 6. (a) Let u = t 2 + 1 J (i = t 2 + 1 =4> du = 2t dt =4> i du = t dt; t = =4> u = 1, t = a/7 => u = 8 t { e + 1) 1/3 dt = £ \ U V3 du = [(i) (|) u 4 / 3 ] ; = (i) ( 8) 4 / 3 - (|) (1) 4 / 3 = f he same substitution as in part (a); t = — y 7 => u = 8, t = =>■ u=l / 3 du=-J^ 8 iu 1/3du =-f ._..,■. .nv. ,.„,„. ....v. ...,,v..,v, u as in part (a); t = -\/7 => u = 8, t = =>■ u=l v< t2+1) dt =J> 1/0 " "' " 7. (a) Let u = 4 + r 2 du = 2r dr =► ^ du = r dr; r = — 1 =>• u = 5, r = 1 =>• u = 5 / — ^ dr = 5 XV du = (b) Use the same substitution as in part (a); r = => u = 4, r = 1 => u = 5 r^ dr = 5 r^- 2 du = 5[-iu-^ = 5(-i(5r 4 )-5(-i(4ri)=I (a) Let u = 1 + v 3 / 2 => du = | v 1 / 2 dv =>■ ^ du = lOy^v dv;v = => u=l,v=l => u p 1 r- n 2 n 2 I lOv/v , ( i n(\ , \ ■■il i _■■ , in r : - ■ - >ii i i n in Jo ft- ,3 /2f dV=j; 2 ^(fdu)=f / 2 U-MU: :- ^7 :■■ ^ "~™ 3 LuJ 1 ~ 3 L2 1 J 3 (b) Use the same substitution as in part (a); v=l => u = 2, v = 4 =>• u = 1 + 4 3 / 2 = 9 du ) = _ 20 r 11 9 _ 20 f 1 P !°V^ dv _ f 9 J_ f20 j\ (l +v 3/2) 2 QV ~ J 2 " 2 V3 9. (a) Let u = x 2 + 1 -r uu-^u* -^ I ^=dx= J 4-du = 2U- 1 3 LuJ 2 20 fl _ 1\ _ _ 20 /_ J7_\ _ 70 3 V9 2^ — 3 I 18) ~ 27 u = 1, x = \/3 =4> u = du = 2x dx =4- 2 du = 4x dx; x = rV2 du = [4u 4 / 2 ] J = 4(4) J / 2 - 4(l) 1 /2 = (b) Use the same substitution ; as in /v/3 n4 -v/3 \/x 2 + l J 4 v/ U 10. (a) Let u = x 4 + 9 part (a); x = — \/ 3 => u = 4, x = \/3 =>- u => du = 4x 3 dx => i du = x 3 dx; x = =>■ u = 9, x = 1 =>• u = 10 /.' 7^ dx = X' 1 "- 1/2 du = [I ( 2 ) ul/2 ] J° = I d0) 1/2 - 2- (9) 1/2 = #=2 ■., v/x'i+9 "'" J9 4" — L4V- J 9 2-"' 2^> 2 (b) Use the same substitution as in part (a); x = — 1 =>■ u = 10, x = =>• u = 9 ^^dx= / i U - 1 / 2 du=- I U - 1 /2 d U= 3 ^^ -l \/x4 + 9 J 10 4 J 9 4 2 11. (a) Let u = 1 - cos 3t =^> du = 3 sin 3t dt ^ \ du = sin 3t dt; t = => u = 0, t = § =^> u = 1 - cos | = 1 £ /6 (l-cos3t)sin3tdt=/ o 1 iudu=[i(f)]^=i(l) 2 -I(0) 2 =i (b) Use the same substitution as in part (a); t = | =£> u=l,t=| => u=l — cos 7r = 2 ■ v ft (1 — cos 3t) sin 3t ■ f-Tft Jx/6 u ^t=/ju^=[i(f)]; : 1(2)^-1(1)2 = i Copy |t (c| 1 Pearson Etoatioo, Inc., publishing as Pearson Addison-Wesle Section 5.6 Substitution and Area Between Curves 33 1 12. (a) Let u = 2 + tan \ => du = \ sec 2 | dt => 2 du = sec 2 | dt; t = =^ => u = 2 + tan (=^) = 1, t = =$> u = 2 J ^ (2 + tan ±) sec 2 \ dt = J u (2 du) = [u 2 ] J = 2 2 - l 2 = 3 (b) Use the same substitution as in part (a); t = -£ => u=l,t=| => u = 3 J"" 2 (2 + tan I) sec 2 \ dt = 2 J u du = [u 2 ]j = 3 2 l 2 = 8 13. (a) Let u = 4 + 3 sin z =4> du = 3 cos z dz =>• J du = cos zdz;z = =>• u = 4, z = 2n =4> u = 4 /A cos *. dz = / 4- (A du) = Jo y 4 + 3 sin z J 4 V" v ' ' (b) Use the same substitution as in part (a); z = — 7r =4> u = 4 + 3 sin (— 7r) = 4, z = ir =>■ u = 4 £ x/4 + 3! dz r^(5 du )=° 14. (a) Let u = 3 + 2 cos w =$■ du = —2 sin w dw =>• — | du = sin w dw; w = — f =>■ u = 3, w = =>• u = 5 dw r» --' 1 du) = ifu- 1 i 5 - ±fl n 15 _^2 (3 + 2cosw) 2 " vv J 3 " V 2 ""7 2 L" J3 2 V5 3 V (b) Use the same substitution as in part (a); w = =>■ u = 5, w = | =>■ u = 3 r „J nw a dw = f V 2 (- | du) = i f u- 2 du = i Jo (3 + 2 cos w) 2 J 5 V 2 / 2 J 15 15. Let u = t 5 + 2t =>■ du = (5t 4 + 2) dt; t = =>■ u = 0, t = 1 =4> u = 3 / o ' y/W+K (5t 4 + 2) dt = JV/ 2 do = [| u 3 / 2 ] J = f (3) 3 / 2 - f (0) 3 / 2 =2^ 16. Letu=l+ v /y =4> du=^;y = l => u = 2, y = 4 =>• u f 4 ' y ^ = f 3 \du= fV 2 du=[-u- 1 ]2 J i 2J^(l + Jyi 2 J2 u2 ^ L J2 2^ 6 17. Let u = cos 261 => du = -2 sin 20 d0 =>• - § du = sin 20 d0; = => u = 1, = | => u = cos 2 (f ) = \ r^ 6 r 1/2 r 1/2 r / _,m i/ 2 J o cos- 3 20 sin 20 d0 = J L u - 3 (-idu) = -lJ | u~ 3 du = [- \ (^)] j 4 (i) 2 4(1)2 4 18. Let u = tan (I) => u = tan | = 1 du = 1 sec 2 (|) d0 => 6 du = sec 2 (§) d0; = tt =>■ u = tan (f ) = -A- . ni*/2 ^t 5 (f) sec 2 (f ; « w =C u " 5(6du) =[ 6 (^)]U=[-* ] 2u<U 1/^/3 3 2(1)4 12 .s/i) 19. Let u = 5 - 4 cos t =4> du = 4 sin t dt =>■ | du = sin t dt; t = =^ u = 5 - 4 cos = 1, t = n =4> u = 5 — 4 cos 7r = 9 J*5 (5-4 cos t) 1 / 4 sin t dt = J^u 1 / 4 (A du) = | J^u 1 / 4 du = [§ (f u 5 / 4 )] J = 9 5 / 4 - 1 = 3 5 / 2 - 1 20. Let u = 1 - sin 2t => du = -2 cos 2t dt =>■ - \ du = cos 2t dt; t = =4> u = 1, t = | => u = /^(l - sin 2t) 3 / 2 cos 2t dt = f- 1 u 3 / 2 du = [- A (§ u 5 / 2 )] J = (- 1 (Of' 2 ) - (- A (l) 5 / 2 ) = \ 21. Letu = 4y-y 2 + 4y 3 + 1 => du = (4 - 2y + 12y 2 ) dy; y = =4> u = 1, y = 1 => u = 4(1) - (l) 2 + 4(1) 3 + 1 J (4y - y 2 + 4 y 3 + 1)- 2 / 3 ( 12y 2 _ 2y + 4) dy = J] u- 2 / 3 du = [3u 4 / 3 ] \ = 3(8) 4 / 3 - 3(1) 4 / 3 = 3 Copffigl (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 332 Chapter 5 Integration 22. Let u = y 3 + 6y 2 - 12y + 9 =4> du = (3y 2 + 12y - 12) dy =>• | du = (y 2 + 4y - 4) dy; y = => u = 9, y = 1 => u = 4 J o '(y 3 + 6y 2 - 12y + 9)" 1/2 (y 2 + 4y - 4) dy = f f \ u- 1 ' 2 du = [I ^u 1 ^)] J = | (4)V2 _ | (9) i/2 = | ( 2 _ 3) _ 2 3 23. Let u = 6» 3 / 2 =>- du = 1 (9 1 / 2 d0 =s> | du = \/9 d0; 9 = => u = 0, 6» = Vtt 2 ^ u = tt J* V "v/0 cos 2 (# 3 / 2 ) dfl = JJcos 2 u (I du) = [ 2 fu , I 3 l2 T 4 2u)] ~~ 3 \2 "^ 4 sin 2tt) - I (0) = f 3 v"; — 3 24. Let u = 1 + \ =^ du = -t~ 2 dt; t = — 1 => u = 0, t = - \ => u = -1 £~'V 2 sin 2 (1 + I) dt = Jf-sin 2 udu =[-(!- 1 sin 2u)] - U - 5 - | sin '(-2)) '0 _ 1 v2 4 sin 0) \ - \ sin 2 25. Let u = 4 - x 2 =4> du = -2x dx => - | du = x dx; x = -2 => u = 0, x = =>■ u = 4, x = 2 => u = A = - J x^4 - x 2 dx + J x^4 - x 2 dx = - J - ± u 1 / 2 du + J - \ u 1 / 2 du = 2 J \ u 1 / 2 du = J u 1 / 2 du = [I u3/2 ] o = I W 3/2 " I <°) 3/2 = f 26. Let u = 1 — cos x =4> du = sin x dx; x = =>- u = 0, x = 7r =>• u = 2 2 f f 1 - cos x) sin x dx = J u du = | 4 I = =f - -, : 2 (i 2 2 27. Let u = 1 + cos x =>■ du = —sin x dx => — du = sin x dx; x = —ir =^ u = 1 + cos (— ir) = 0, x = =>■ u = 1 + cos = 2 A = - J_ 3 (sin x) a/1 +cosx dx = - J 3U 1 / 2 (-du) = 3 J u 1 / 2 du = [2u 3 / 2 ] 2 = 2(2) 3 / 2 - 2(0) 3 / 2 = 2 5 / 2 28. Let u = 7r + 7r sin x =>• du = 7r cos x dx =>■ ^ du = cos x dx; x = — | =>■ u = 7r + 7r sin (— f ) = 0, x = =>• u = 7r | (cos x) (sin (7T + 7r sin x)) dx = 2 I | (sin u) (- du) = I sin u du = [—cos u]q = (—cos jf) — (—cos 0) = 2 1 — cos 2x . 29. For the sketch given, a = 0, b = n; f(x) — g(x) = 1 — cos x = sin x — A= J**iL = »!M dx= I < £ ,r (1 _ COS 2 X )dx=i[x-^];=i[(7r-0)-(0-0)]=| 30. For the sketch given, a 2f sec 2 t + 4 sin 2 t dt /lr/3 -ir/ 3 I sec 2 1 dt + 2 | */ — 7r/3 «./ — rr f ; f(t) - g(t) = \ sec 2 1 - (-4 sin 2 1) = \ sec 2 1 + 4 sin 2 t /tt/3 /-»7r/3 P 71 "/^ p 71 "/^ sec 2 1 dt + 4 sin 2 1 dt = \\ sec 2 1 dt + 4 (1 " c o 0s2t) dt -tt/3 J-jr/3 2 J-tt/1 J -tt/: Kfi ■k/3 ,, , ^sec 2 1 dt + 2j_ n (l - cos 2t) dt = \ [tan t]_J /3 + 2[t - ^j"'" = v /3+4-f- v / 3=f 31. For the sketch given, a = -2, b = 2; f(x) - g(x) = 2x 2 - (x 4 - 2x 2 ) = 4x 2 - x 4 ; A = f ^ (4x 2 - x 4 ) dx = [ 4x 3 _ x^ 3 5 '32 _ 32 \ 32 3 32 M _ 64 64 _ 320-192 _ 128 ! )] Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle Section 5.6 Substitution and Area Between Curves 333 32. For the sketch given, c = 0, d = 1; f(y) — g(y) = y 2 — y 3 ; (1-0) _ (1-0) _ 1 _ 1 _ J_ 3 4 — 3 4 — 12 33. For the sketch given, c = 0, d = 1; f(y) - g(y) = (12y 2 - 12y 3 ) - (2y 2 - 2y) = 10y 2 - 12y 3 + 2y; A=/ o '(10y 2 -12 y 3 + 2 y )d y = / o I 10 y 2 dy-/ o 1 12 y 3 dy + / o '2ydy=[fy3];~[fy4]; + [|y 2 ]; = (f -0)- (3-0) + (1-0) = | 34. For the sketch given, a = -1, b = 1; f(x) - g(x) = x 2 - (-2x 4 ) = x 2 + 2x 4 ; A J> + 2x 4 )dx=[f + f]^ = (i + f) -[-i + (- 2 )] 3^5 10+12 _ 22 15 ~ 15 35. We want the area between the line y = 1, < x < 2, and the curve y = \, minus the area of a triangle (formed by y = x and y = 1) with base 1 and height 1. Thus, A I'M) dx-i(l)(l) = (2- J M-i=2- 2 -- i = 5 K* 12 J 2 A 3 2 6 36. We want the area between the x-axis and the curve y = x 2 , < x < 1 plus the area of a triangle (formed by x = 1, x + y = 2, and the x-axis) with base 1 and height 1 . Thus, A Jo dx+i(l)(l) 1 _ 1 , 1 _ 5 2 ~~ 3 "r 2 ~~ 6 37. AREA = Al + A2 Al: For the sketch given, a = — 3 and we find b by solving the equations y = x 2 — 4 and y 2x simultaneously for x: x 2 - 4 = -x 2 - 2x =4> 2x 2 + 2x - 4 = => 2(x + 2)(x - 1) => x = -2 or x = 1 so b = -2: f(x) - g(x) = (x 2 - 4) - (-x 2 - 2x) = 2x 2 + 2x - 4 => Al /> x 2 + 2x - 4) dx ¥ + ¥-4*i v , 16 ' 4+8) -(-18 + 9+ 12) = 9- 16 - " A2: For the sketch given, a = -2 and b = 1: f(x) - g(x) = (-x 2 - 2x) - (x 2 - 4) = -2x 2 - 2x + 4 3 — 3 2 _ o v \ _ f v 2 _ A\ _ _o v 2 => A2 = - J_ n (2x 2 + 2x - 4) dx = - Uf + x 2 - 4x1 = - (§ + 1 - 4) ;-f+ 4 - 5-1+4-^+4 + 8 = 9; 3S Therefore, AREA = Al+A2=-^+9- , 38. AREA = Al + A2 Al: For the sketch given, a = -2 and b = 0: f(x) - g(x) = (2x 3 - x 2 - 5x) - (-x 2 + 3x) = 2x 3 - 8x Al £(2x3 8x) dx 2x 4 8x 2 0-(8- 16) = 8; A2: For the sketch given, a = and b = 2: f(x) - g(x) = (-x 2 + 3x) - (2x 3 - x 2 - 5x) = 8x - 2x 3 => A2 = j g (8x - 2x 3 ) dx = fe - ^1 = (16 - 8) = 8; Therefore, AREA = Al +A2 16 39. AREA = Al + A2 + A3 Al: For the sketch given, a = — 2 and b = — 1: f(x) — g(x) = (— x + 2) — (4 — x 2 ) = x 2 — x — 2 =* Al = £ 1 (x 2 -x-2)dx=[f-f-2x]^=(-3-i+2)-(-f- 4 +4) A2: For the sketch given, a = - 1 and b = 2: f(x) - g(x) = (4 - x 2 ) - (-x + 2) = - (x 2 - x - 2) 7 _ 1 _ 14-3 _ n. 3 2 6 6 ' A2 ■/: < x 2 - x - 2) dx 2x '8 _ 4 v3 2 4) l-i+2) -3 + 8- ^ - 2'' Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 334 Chapter 5 Integration A3: For the sketch given, a = 2 and b = 3: f(x) - g(x) = (-x + 2) - (4 - x 2 ) = x 2 - x - 2 =>. A3 = J (x 2 - x - 2) dx Therefore, AREA = Al + A2 + A3 2x 6 + (9 '27 6) 4) 49 6 40. AREA = Al + A2 + A3 Al: For the sketch given, a = -2 and b = 0: f(x) - g(x) = ( y - xj - § = y - | x = | (x 3 - 4x) Al £ (* 3 4x) dx 1 Ui 3 4 2x< -2 3 (4- 8)= |; A2: For the sketch given, a = and we find b by solving the equations y x 3 „ x _ , x 3 4 forx: j - x = | =>• y - j x = f(x)-g(x) = f-(f -x) =-|(x | (x - 2)(x + 2) = => x '' -4x) =>■ A2 = - i J (x 3 - 4x) dx = ± Jj (4x - x 3 ) = 1 |2x j — x and y = f simultaneously —2, x = 0, or x = 2 so b = 2: 2 4) A3: For the sketch given, a = 2 and b = 3: f(x) - g(x) = ( y - xj - | = | (x 3 - 4x) =► A3= 3 /;(x 3 -4x)dx=I^-2x 2 i;= 3 [( T -2.9)-(f-8)] I (f - 14) Therefore, AREA = Al + A2 + A3 25 12 32+25 19 4 25. 12' 41. a= -2, b = 2; f(x) - g(x) = 2 (x 2 - 2) = 4 - x 2 A= J_ 7 (4-x 2 )dx= Ux '24 32 3 42. a= -1, b = 3; f(x) - g(x) = (2x - x 2 ) - (-3) =-. A = f (2x - x 2 + 3) dx = = (9-f + 9)-(l + i-3) 2x 11 3x 32 3 y=2x-x 2 43. a = 0, b = 2; f(x) - g(x) = 8x 2 16 32 5 = /> 80-32 _ 48 5 — 5 x 4 )dx 44. Limits of integration: x 2 — 2x = x =4> i => x(x - 3) = => a = and b = 3; f(x) - g(x) = x - (x 2 - 2x) = 3x - x 2 3x > A = ¥-9 J> x - x 2 ) dx = I ¥ - 4 27-18 _ 9 2 2 Copyright (c) 2006 Pearson Education 45. Limits of integration: x 2 = — x 2 + 4x =>• 2x 2 — 4x = => 2x(x - 2) = =4> a = and b = 2; f(x) - g(x) = (-x 2 + 4x) - x 2 = -2x 2 + 4x J (-2x 2 + 4x) dx -32 + 48 _ 8 Section 5.6 Substitution and Area Between Curves y 335 A -2x 3 | 4x^ 3 + 2 16 , 16 3 + 2 46. Limits of integration: 7 - 2x 2 = x 2 + 4 => 3x 2 - 3 = =>• 3(x- l)(x+ 1) = => a=-landb=l; f(x) - g(x) = (7 - 2x 2 ) - (x 2 + 4) = 3 - 3x 2 => A = J(3 - 3x 2 ) dx = 3 [x - f ] * = 3[(l-|)-(-H-I)]=6(l)=4 _1 y=7-2x2 47. Limits of integration: x 4 — 4x 2 + 4 = x 2 => x 4 - 5x 2 + 4 = => (x 2 - 4) (x 2 - 1) = => (x + 2)(x - 2)(x + l)(x - 1) = => x = -2, -1, 1, 2; f(x) - g(x) = (x 4 - 4x 2 + 4) - x 2 = x 4 - 5x 2 + 4 and g(x) - f(x) = x 2 - (x 4 - 4x 2 + 4) = -x 4 + 5x 2 - 4 => A = J (-x 4 + 5x 2 - 4)dx + J (x 4 - 5x 2 + 4)dx + J f (-x 4 + 5x 2 - 4)dx - :i 4x| ^ + [f - f + 4x] ^ + [=* + f - 4x] \ (i-| + 4)-(f-f + 8) + (I-f + 4)-(-I + f-4) + 60 , 60 _ 300-180 _ o ~ T + T - vT~ ~ 8 32 , 40 5 + 3 1 + ^ 5^3 4) 48. Limits of integration: xa/ a 2 — x 2 = =^ x = or \/a 2 - x 2 =0 =4> x = or a 2 - x 2 = =>■ x = -a, 0, a; A= / -xy 7 a 2 - x 2 dx + / xy 7 a 2 - x 2 dx 1 \2 /„2 „2n3/2 , L ., ( a-- -xT i(a 2 r- -I(a^) I [| (a 2 - x 2 ) 3/i -x, x < x i 6 5 T 5 49. Limits of integration: y V x, x > 5y = x + 6ory= | + f; for x < 0: \J — x =>■ 5^^ = x + 6 =^ 25(-x) = x 2 + 12x + 36 => x 2 + 37x + 36 = => (x + l)(x + 36) = =4> x = — 1, —36 (but x = —36 is not a solution); for x > 0: 5y/H = x + 6 => 25x = x 2 + 12x + 36 => x 2 - 13x + 36 = => (x - 4)(x - 9) = => x = 4, 9; there are three intersection points and 6 and A j: x + 6 ^x+x^-v^+r^ y = V :: x 4 s ) dx Cfiojt (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 336 Chapter 5 Integration (x + 6) 2 , 2 10 + 1 (~*) 3/2 ] \ +[^f-l * 3/2 ] I + [f ^ " ^f] ' '36 _ 25 _ 2\ , / 100 _ 2 4 3/2 _ 36 , ft \ , (2 q 3/2 _ 225 _ 2 4 3/2 , K>0\ vlO 10 3) ' \ 10 3" H 10 1" U J ' I 3 ' * 10 3" H "I" 10 / 50 , 20 5 10 ' 3 3 50. Limits of integration: x 2 - 4, x < -2 or x > 2 4 - x 2 , -2 < x < 2 for x < -2 and x > 2: x 2 - 4 = f + 4 => 2x 2 - 8 = x 2 + 8 = for -2 < x < 2: 4 - x 2 => x 2 = => x = 0; by symmetry of the graph, j/ = i 2 /2 + 4 = 2(|-0)+2(32-f-16 + |) 40-56-64 tu 3 — 3 51. Limits of integration: c = and d = 3; f(Y) - g(y) = 2y 2 - = 2y 2 / V dy 2-9= 18 1 — x 52. Limits of integration: y 2 = y + 2 => (y + l)(y — 2) = - 1 and d -.2 2; f(y) - g(y) = (y + 2) - y 2 A J> + 4- 2 - y 2 ) dy = ^+2y 6-f- 53. Limits of integration: 4x = y 2 — 4 and 4x = 16 + y => y 2 -4=16 + y => y 2 -y-20 = => (y - 5)(y + 4) = => c = -4 and d = 5; f(y)-g(y)= C-T) - (^) = =^T^ 80) A = = U_ 4 (-y 2 + y+20)d y if- ^+20y] 5 4 u- H- f + f + 100)-i(f + f f + 1 + 180) = ^ -y=16 Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 54. Limits of integration: x = y 2 and x = 3 — 2y 2 => y 2 = 3 - 2y 2 => 3y 2 = 3 => 3(y - l)(y + 1) = => c = -1 and d = 1; f(y) - g(y) = (3 - 2y 2 ) - y 2 = 3 - 3y 2 = 3 (1 - y 2 ) => A = 3 fj\ - y 2 ) dy Section 5.6 Substitution and Area Between Curves 337 y 3-2(l-I)=4 3(1-1) -3(-l + I) x+2y2=3 x-y2=0 55. Limits of integration: x = — y 2 and x = 2 — 3y 2 => -y 2 = 2 - 3y 2 =4> 2y 2 - 2 = => 2(y - l)(y + 1) = => c = -1 and d = 1; f(Y) - g(Y) = (2 - 3y 2 ) - (-y 2 ) = 2 - 2y 2 = 2 (1 - y 2 ) => A = 2/_ 1 | (l~y 2 )dy = 2[ = 2(l-I)-2(-l + i)=4 y-i 2\ _ 8 3 ^ 3 x+3y 2 =2 56. Limits of integration: x = y 2 / 3 and x = 2 — y 4 => y 2 / 3 = 2 - y 4 => c = -1 and d = 1; f(y) - g(y) = (2 - y 4 ) - y 2/3 => A=/' i (2-y 4 -y 2 / 3 ) dy = [^y-i-h 5/3 }\ = (2-I-|)-(-2+I + |) 2(2 1 3\ _ 12 57. Limits of integration: x = y 2 — 1 and x = |y| \/l — y 2 =>. y 2 - 1 = |y| ^/1-y 2 => y 4 - 2y 2 + 1 = y 2 (1 - y 2 ) =^ y 4 - 2y 2 + 1 = y 2 - y 4 => 2y 4 - 3y 2 + 1 = => (2y 2 - 1) (y 2 - 1) = => 2y 2 - 1 = or y 2 - 1 = v 2 = l „, „2 y 2 ory 2 =l => y= ± ^ or y = ± 1 . ± /I Substitution shows that — 5j— are not solutions => y = ±1; for -1 < y < 0, f(x) - g(x) = -y y/l - y 2 - (y 2 - 1) = 1— y 2 — y (1 — y 2 ) , and by symmetry of the graph, A = 2£[l-y 2 -y(l-y 2 ) 1/2 ]dy = 2/_ i (l-y 2 )dy-2£° i y(l- y 2 ) 1/2 dy 2y- '1\ |2(l-y 2 ) 3/2 i x =1 y I \/i -y 2 2[(0-0)-(-l + i)] + (|-0)=2 Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 338 Chapter 5 Integration 58. AREA = Al + A2 Limits of integration: x = 2y and x = y 3 — y 2 => y 3 _ y 2 = 2 y =>. y (y 2 - y - 2) = y(y + l)(y - 2) = ^ y = -1,0,2: for - 1 < y < 0, f(y) - g(y) = y 3 - y 2 - 2y =* Al = /_" i (y 3 -y 2 -2y)dy=[^-f = o-(i + J-i) = & for < y < 2, f(y) - g(y) = 2y - y 3 + y 2 => A2 = / o 2 (2y-" 3 ' •■-> Therefore, Al+A2=-^ + f = g r y J + y'J dy y 2 -^ 59. Limits of integration: y = — 4x 2 + 4 and y = x 4 — 1 1 = -4x 2 +4 => x 4 + 4x 2 => (x 2 + 5)(x- l)(x+ 1) = => a= -1 andb= 1; f(x) - g(x) = -4x 2 + 4 - x 4 + 1 = -4x 2 - x 4 + 5 A Jj- 4 _ 1 3 5 -4x 2 - 5)- 5)dx 4x 3 3 + 5x |-5)=2(-|-| + 5) -l 104 15 60. Limits of integration: y = x 3 and y = 3x 2 — 4 =^ x 3 - 3x 2 + 4 = => (x 2 - x - 2) (x - 2) = => (x + l)(x - 2) 2 = => a = -1 and b = 2; f(x) - g(x) = x 3 - (3x 2 - 4) = x 3 - 3x 2 + 4 2 •4x1 A = j (x 3 - 3x 2 + 4) dx = h )-a+i-4)=f 3xf 4 3 16 _ 24 4 3 61. Limits of integration: x = 4 — 4y 2 and x = 1 — y 4 => 4 - 4y 2 = 1 - y 4 => y 4 - 4y 2 + 3 = =>• (y-v^) (y + v/3)(y-i)(y+i) = o =* c = -i and d = 1 since x > 0; f(y) - g(y) = (4 - 4y 2 ) - (1 - y 4 ) = 3 - 4y 2 + y 4 => A = J__(3 - 4y 2 + y 4 ) dy 3y 4y 3 2(3-f + i) 56 15 x =4-4y2 x=1-y 4 62. Limits of integration: x = 3 — y 2 and x y~ 4 3 = => | (y - 2)(y + 2) = -2 and d 2;f(y)-g(y) = (3-y 2 )-(^) 3(l-fl ^A = 3£(l-£)dy = 3[; 3[(2-&)-(-2+£)]=3(4 y-i \l) 12-4 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 5.6 Substitution and Area Between Curves 339 63. a = 0, b = 7r; f(x) — g(x) = 2 sin x — sin 2x => A = J (2 sin x - sin 2x) dx = [-2 cos x + ^] £ = [-2(-l)+i]-(-2-l + i)=4 y=2sin x 64. a = — |, b = |; f(x) — g(x) = 8 cos x — sec x (8 cos x — sec 2 x) dx = [8 sin x — tan x] _ = (8-f-V3)-(-8-#- /3 6\/3 y=8cosx (secx) 2 * x 65. a = -1, b = 1; f(x) - g(x) = (1 - x 2 ) - cos A: I J 1 — X — COS I i)]dx 2(i-i; 4 _ 4 3 7T y=1-x2 66. A = Al + A2 ai = — 1, bi = and a 2 = 0, b2 = 1; fi(x) - gi(x) = x - sin (f ) and f 2 (x) - g 2 (x) = sin (f ) - x =>• by symmetry about the origin, Ai + A 2 = 2Ai => A = 2 J [sin (f ) - x] dx , =2[(-f"0-i)-(-f-l-0)] 2 I- 2 cos (^ V Z7T y=sin(*x/2) — i — x 67. a = — f , b = |; f(x) - g(x) = sec 2 x - tan 2 x / 7r/4 (sec 2 x — tan 2 x) dx - /. [sec 2 x — (sec 2 x — 1)] dx ?r/4 L /i-dx [xl;, 7T/4 / -7r/4 ,7T/4 68. c = - |, d = |; f(y) - g(y) = tan 2 y - (- tan 2 y) = 2 tan 2 y = 2(sec 2 y- 1) => A= I 2(sec 2 y-l)dy *^ -7r/4 W 4 4 - 7T 2[tany-y]! / 7r 4 /4 =2[(l 4fl — ? ^/4 -jt/4' •i + !)] x=(tany) 2 i — x Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 340 Chapter 5 Integration 69. c = 0, d = |; f(y) — g(y) = 3 sin y^/cos y — = 3 sin y^/cos y => A = 3 J" sin y ycos^ dy = -3 [| (cos y) 3 / 2 ] ^ = -2(0 - 1) = 2 70. a = -1, b = 1; f(x) - g(x) = sec 2 (^ 3 V3] dX 3 -tan I 6^3 t* 4/3 ]-i 1- x 71. A = Aj +A 2 Limits of integration: x = y 3 and x = y => y 3 - y = =>• y(y - l)(y + 1) = and c 2 = 0, d 2 y = y° ci = — 1, di f2(y) - g2(y) = y Ai + A 2 = 2A 2 : i;fi(y)-gi(y) = y -y and — y 3 => by symmetry about the origin, * A = 2f'(y-y 3 )dy = n 2(1 l 72. A = Ai + A 2 Limits of integration: y = x 3 and y = x 5 x 3 (x - l)(x + 1) = X' = x u » ai = -1, bi and a 2 = 0, b 2 = 1; f x (x) - g x (x) x 5 and f2(x) - g 2 (x) = J Ai + A 2 = 2A 2 by symmetry about the origin, A '/.V dx 2 a -i# 73. A = Ai + A 2 Limits of integration: y = x and y = \ =>- x=^,x^0 =4> x 3 = 1 => x = 1 , fi(x) - gi(x) = x - = x =, Al = /Jx dx = [f ] * = I; f 2 (x) - g 2 (x) = 4, - =*- 2 ^ A 2 = jV 2 dx=[^] 2 = - 1 A = Ai + A 2 2. 1/ = 1/x Copffigl (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 5.6 Substitution and Area Between Curves 341 74. Limits of integration: sin x = cos x => and b = | ; f(x) — g(x) = cos x — sin x >?r/4 f => a = pTT/4 A = / (cos x — sin x) dx = [sin x + cos x] u tt/4 V2 , V2 -(0 + 1) = a/2-1 75. (a) The coordinates of the points of intersection of the line and parabola are c = x 2 =>- x = ± yfc and y = c (b) f(y) — g(y) = y/y — (— -t/y) — 2,/y => the area of the lower section is, A L = I [f(y) — g(y)] dy = 2 J y/y dy = 2 [| y 3 / 2 ] C Q = f c 3 / 2 . The area of the entire shaded region can be found by setting c = 4: A = ft) 4 3//2 , , , _ ^r = T • Since we want c to divide the region into subsections of equal area we have A = 2A L 32 3 2(fc 3 / 2 ) 12/3 (c) f(x) - g(x) = c - x 2 => A L = J ^ [f(x) - g(x)] dx = / (c - x 2 ) dx = [ex - 2, ,1, - .... vl ^ =2 |c 3 / 2 - c3: = I c 3 ' 2 . Again, the area of the whole shaded region can be found by setting c = 4 =>■ A condition A = 2A L , we get | c 3/2 = y => c = 4 2/3 as in part (b). y . From the 76. (a) Limits of integration: y = 3 — x 2 and y = — 1 -1 a -2 and b f(x)-g(x)=(3-x 2 )-(-l) = 4-x =$► A = f (4 - x 2 ) dx = Ux - 16 -2 16 — 32 3 — 3 (b) Limits of integration: let x = in y = 3 — x 2 => y = 3; f(y) - g(y) = v / 3 T ~y - (- x/3^7) = 2(3 - y) 1 / 2 => A = 2 J (3 - y) 1/2 dy = -2 J (3 - y) 1/2 (-l) dy = (-2) !/ = -! 2(3 -y) 3 3 i)(8) 32 3 [0 - (3 + l) 3 / 2 ] 77. Limits of integration: y = 1 + ^Jx and y = -j- => l + v /^=2^ x _^ ^ v /^ + x = 2 => x = (2-x) 2 => x = 4 - 4x + x 2 =>• x 2 -5x + 4 = =4> (x - 4)(x - 1) = => x=l,4 (but x = 4 does not satisfy the equation); y = A- and y = | =>• -j- = | x^/x 64 Therefore, AREA = Aj + A 2 : fi(x) - gl (x) = (l ,1/2 )-! * - 1: .1/2 dx 2-8/3 y = l + v/5 ;i + i-i)-o g;f 2 (x)-g 2 (x) = 2x- 1 /2_| A 2 = /"(2X- 1 / 2 - |) dx = [4xV 2 - £] (4-2 -(4-|)=4-f f ; Therefore, AREA = A : + A 2 = g n _ 37+51 8 24 88 24 Cfiojt (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 342 Chapter 5 Integration 78. Limits of integration: (y — l) 2 = 3 — y => y 2 — 2y + 1 = 3 - y => y 2 - y - 2 = => (y- 2)(y + 1) = =4> y = 2 since y > 0; also, 2,/y = 3 — y => 4y = 9 - 6y + y 2 => y 2 - lOy + 9 = =$■ (y — 9)(y — 1) = =>• y=l since y = 9 does not satisfy the equation; AREA = Ai + A 2 fi(y)-gi(y) = 2 v ^-o = 2y 1 / 2 => A 1 = 2 / o 'y 1/2 dy = 2 [^] * = f ; f 2 (y) - g 2 (y) = (3 - y) - (y - l) 2 =* A 2 = / 2 [3 - y - (y - l) 2 ] dy = [3y - \ y 2 - i (y - l) 3 ] \ = (6 - 2 - |) - (3 - \ + 0) = 1 - § + 1 Therefore, Aj + A 2 _4,7_15_5 v 2 -3 + 6-^-2 79. Area between parabola and y = a 2 : A = 2 I (a 2 — x 2 ) Area of triangle AOC: \ (2a) (a 2 ) = a 3 ; limit of ratio = lim dx = 2 [a 2 x - 1 x 3 ] a Q = 2 (a 3 - f ) 0=f; | which is independent of a. nb nb r*b r*b nb 80. A = J 2f(x) dx - J f(x) dx = 2 J f(x) dx - J f(x) dx = J f(x) dx = 4 81. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the region's upper and lower bounding curves at x = 0. The area of the shaded region is actually pO pl pl) pi A = J [-x - (x)] dx + J [x - (-x)] dx = J -2x dx + J 2x dx = 2. 82. It is sometimes true. It is true if f(x) > g(x) for all x between a and b. Otherwise it is false. If the graph of f lies below the graph of g for a portion of the interval of integration, the integral over that portion will be negative and the integral over [a, b] will be less than the area between the curves (see Exercise 53). 83. Let u = 2x => du = 2 dx => | du = dx; x=l =>• u = 2, x = 3 =>• u = 6 J §2^5 dx = J^ stan (i du ) = J a» du = [F(u)] I = F(6) - F(2) 84. Let u = 1 - x =>• du = -dx =4> -du = dx;x = => u=l,x=l =4> u = pl p0 pi) pi pi J f(l-x)dx=J f(u)(-du) = -J f(u)du = J f(u)du=J f(x) dx 85. (a) Let u = -x =^ du = - dx; x = -1 =4> u = 1, x = =>■ u = /0 n0 pa no pl f(x)dx=J f(-u)(-du) = J -f(u)(-du) = J f(u)du=-J f(u) du = -3 (b) Letu = -x => du = - dx; x = -1 =>- u = 1, x = => u = p0 pl) pO pl feven => f(-x) = f(x). Then J f(x) dx = J f(-u)(- du) = - J f(u) du = J f(u) du = 3 86. (a) Consider J f(x) dx when f is odd. Let u = — x => du = — dx =>• — du = dx and x = — a =>• u = a and x = /0 pl) pO pa pa f(x) dx = J -f(-u) du = J f(u) du = - J f(u) du = - J f(x) dx. Thus J f(x) dx = J f(x) dx + J f(x) dx = - J f(x) dx + J f(x) dx = 0. Cop # (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle ->-/2 t -Ttd r (b) J sin x dx = [—cos x] ' , cos n 7T/2 -ir/2 Section 5.6 Substitution and Area Between Curves 343 s(f)+cos(-f) =0 + = 0. 87. Let u = a — x =>- du=— dx; x = =4* u = a, x = a =4> u = T _ P' f(x)dx _ P f(a-u) , .x _ P' f(a-u)du _ P" f(a-x) dx Jo «x)+f(a-x) J a f(a-u)+f(u) ^ UU ' J f(u)+f(a-u) J f(x)+f(a-x -x) Therefore, 21 = a =4> I 2 ' Let u = S =>. du = - S dt => - — du = 7 dt => - - du = 7 dt; t = x => u = y, t = xy t v- xy t u t ' j •> j ("■y pi pi py py I I dt = / - i du = - 1 du = / i du = I - dt Jx t Jy U J U J| U J| t u = 1. Therefore, Let u = x + c =4> du = dx; x = a — c => u = a, x = b — c =>• u = b f»b— c r»b nb I f(x + c) dx = I f(u) du = I f(x) dx l/ a— c *J a «-/ a 90. (a) f(x+1)-(x + 1) (b) y f(x)-slnx (c) t-x f(x + |).sin(x + |) 91-94. Example CAS commands: Maple : f := x -> x A 3/3-x A 2/2-2*x+l/3; g := x -> x-1; plot( [f(x),g(x)], x=-5..5, legend=["y = f(x)","y = g(x)"], title="#91(a) (Section 5.6)" ); ql:=[-5,-2, 1,4]; # (b) q2 := [seq( fsolve( f(x)=g(x), x=ql[i]..ql[i+l] ), i=l..nops(ql)-l )]; for i from 1 to nops(q2)-l do # (c) area[i] := int( abs(f(x)-g(x)),x=q2[i]..q2[i+l] ); end do; add( area[i], i=l..nops(q2)-l ); # (d) Mathematica : (assigned functions may vary) Clear[x, f, g] f[x_] = x 2 Cos[x] g[x_] = x 3 - x Plot[{f[x],g[x]}, {x, -2,2}] After examining the plots, the initial guesses for FindRoot can be determined. pts = x/.Map[FindRoot[f[x]==g[x],{x, #)]&, {-1, 0, 1}] il=NIntegrate[f[x] - g[x], {x, pts[[l]], pts[[2]]}] i2=NIntegrate[f[x] - g[x], {x, pts[[2]], pts[[3]]}] il +i2 Copffigl (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle 344 Chapter 5 Integration CHAPTER 5 PRACTICE EXERCISES 1. (a) Each time subinterval is of length At = 0.4 sec. The distance traveled over each subinterval, using the midpoint rule, is Ah = \ (v ; + v i+1 ) At, where v ; is the velocity at the left endpoint and v i+1 the velocity at the right endpoint of the subinterval. We then add Ah to the height attained so far at the left endpoint v, to arrive at the height associated with velocity v i+1 at the right endpoint. Using this methodology we build the following table based on the figure in the text: t (sec) 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0 v (fps) 10 25 55 100 190 180 165 150 140 130 115 105 90 76 65 h(ft) 2 9 25 56 114 188 257 320 378 432 481 525 564 592 620.2 t (sec) 6.4 6.8 7.2 7.6 8.0 v (fps) 50 37 25 12 h(ft) 643.2 660.6 672 679.4 681.8 NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph. Remember that some shifting of the graph occurs in the printing process. The total height attained is about 680 ft. (b) The graph is based on the table in part (a). h(i eet) , > 700 600 500 400 300 200 100 2 4 6 8 * t (sec) 2. (a) Each time subinterval is of length At = 1 sec. The distance traveled over each subinterval, using the midpoint rule, is As = I (vj + v i+1 ) At, where v ; is the velocity at the left, and v i+1 the velocity at the right, endpoint of the subinterval. We then add As to the distance attained so far at the left endpoint v ; to arrive at the distance associated with velocity v i+1 at the right endpoint. Using this methodology we build the table given below based on the figure in the text, obtaining approximately 26 m for the total distance traveled: t (sec) 1 2 3 4 5 6 7 8 9 10 v (m/sec) 0.5 1.2 2 3.4 4.5 4.8 4.5 3.5 2 s(m) 0.25 1.1 2.7 5.4 9.35 14 18.65 22.65 25.4 26.4 (b) The graph shows the distance traveled by the moving body as a function of time for < t < 10. 10 , 10 (a) E | = \ E a t = \ (-2) = - \ k = 1 k = 1 10 10 (b) E (b k -3a k )= E K- k= 1 k= 1 10 -3 E a k= 1 10 10 10 (c) E (a k + b k -l)= E a t + E b t - k= 1 k= 1 k= 1 10 - E i = k= 1 = -2 + 25-(l)(10) = 13 25 - 3(-2) = 31 Copyright (c) 2006 Pearson Education Chapter 5 Practice Exercises 345 (d) £ (§-W)= £ |- £ b k = |(10)-25 = k= 1 k= 1 k = 1 20 20 4. (a) £ 3a k = 3 £ a k = 3(0) = k= 1 k= 1 20 20 20 (c) £ (3-^) = £ |-7 S b, = i (20) - f (7) k=l k=l k=l 20 20 20 (d) £ (a* -2)= £ a t - £ 2 = - 2(20) = -40 k= 1 k= 1 k= 1 20 20 20 (b) £ (a k + b t )= £ a k + £ b k = + 7 = 7 k= 1 k= 1 k= 1 5. Let u = 2x - 1 =4> du = 2 dx =>- \ du = dx; x = 1 => u = 1, x = 5 => u = 9 £(2x - 1)+ 2 dx = £u+ 2 (i du) = [U 1 / 2 ] J = 3 - 1 = 2 6. Let u = x 2 - 1 => du = 2x dx =4> i du = x dx; x = 1 => u = 0, x = 3 => u = 8 £x (x 2 - 1) 1/3 dx = JV/ 3 (1 du) = [| u 4 / 3 ] J = I (16 - 0) = 6 7. Let u = § => 2du = dx;x = -7r => U = - | , x = => u = J cos (!) dx = J (cos u)(2 du) = [2 sin u]^ /2 = 2 sin - 2 sin (- §) = 2(0 - (- 1)) = 2 Let u = sin x =>■ du = cos x dx; x = => u = 0, x = f =4- u=l Jo (sin x)(cos x) dx Jo du 9. (a (c (c 10. (a (c (c J f(x) dx = | J 3 f(x) dx = 1 (12) = 4 (b) J f(x) dx = J f(x) dx - J f(x) dx = 6 - 4 = 2 /; 2 g ( x) dx = - £ g(x) dx = - / g(x) dx (d) J (-■ 7r g(x)) dx = -7r J g(x) dx = -7r(2) = -2tt £(«+lW) dx= l£f(x)dx+l£g(x)dx=I(6) + i (2) / o g(x) dx = i J o 7 g(x) dx = i (7) = 1 (b) £g(x) dx = J o g(x) dx - f o g(x) dx = 1 - 2 = - 1 £f ( x)dx = -£f( x) dx = — IT (d) f \/2 f(x) dx = \fl \ f(x) dx = V^OO = tt^ */ t/ f [g(x) - 3 f(x)] dx = f g(x) dx - 3 f f(x) dx = 1 - 3tt *J */ t/ o 11. x 2 -4x + 3 = => (x-3)(x- 1) = => x = 3 orx= 1; Area = J (x 2 - 4x + 3) dx - J (x 2 - 4x + 3) dx f - 2x 2 + 3x1 - [y - 2x 2 + 3x] (^ -2(1) 2 + 3(1)) -O] - [(f-2(3) 2 + 3(3))-(f-2(l) 2 + 3(l))] (l + i)- [o- (1 + 01 = | f (x) = x - 4x + 3 Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 12. l-^-=0^4-x 2 -0^x=±2 346 Chapter 5 Integration 1-^=0 => 4 - x 2 - => Area =£( i -f)* t -/ 4 3 ( i -f)* t (3-fi) 2- £ ^ 12 n _ 4\ = B U 3^ 4 f(x) = 1 - (x Z /4) 13. 5 - 5x 2 / 3 = => 1 - x 2 / 3 = => x = ± 1; Area = J (5 - 5x 2 / 3 ) dx - J (5 - 5x 2 / 3 ) dx = [5x - 3x 5 / 3 ] ^ - [5x - 3x 5 / 3 ] I = [(5(1) - 3(1) 5 / 3 ) - (5(-l) - 3(-l) 5 / 3 )] - [(5(8)-3(8) 5 / 3 )-(5(l)-3(l) 5 / 3 )] = [2 - (-2)] - [(40 - 96) - 2] = 62 J f(x) = 5 - 5X 273 \2 4 6 8 -5 \. -10 ^N. -15 ^-s 14. 1 - yx = => x= 1; Area = £ (l - ^x) dx - f*(l ^/i) dx = [ X _| X 3/ 2] l_ [x _| x 3/ 2] 4 = [(1 _ | (1)3/2) _ ] „ [(4 _ 2 (4) 3/2) _ (1 _ 2 (1) 3/ 2) ] = i-[(4-¥)-J]=2 15. f(x) = x, g(x) = 4j, a = 1, b = 2 => A = f [f(x) - g(x)] dx =j; 2 (x-i)dx=[f + ^=(i + i)-(i+i) = i 16. f(x) = x, g(x) = -i=, a =l,b = 2 => A= f [f(x) - g(x)] dx J!V*) dx =[^2v^ f-2^2 :i-2) 7-4V2 f(x) = 1 -Vx 5 f 2 1- y=x^ tllli y=VV X i 2 17. f(x) = (1 - ,/x) 2 , g(x) = 0, a = 0, b = 1 ^ A = J[ [f(x) - g(x)] dx = J q (l - y^) 2 dx = £ (l - 2,/x" + x) dx J 'l-2x 1 / 2 + x)dx=: |x-fx 3 / 2 + ^ 4 + I = I 3 > 2 6 (6-8 + 3) r*b r»\ nl 18. f(x) = (1 - x 3 ) 2 , g(x) = 0, a = 0, b = 1 => A = J [f(x) - g(x)] dx = J (1 - x 3 ) 2 dx = J (1 - 2x 3 + x e dx X I X 1 ' 7 | 1 2 ' 7 14 Copffigl (c| 1 Pearson Education, Ire, publishing as Pearson Addison-Wesle Chapter 5 Practice Exercises 347 19. f(y) = 2y 2 , g(y) = 0, c = 0, d = 3 =>• A = J [f(y) - g(y)] dy = J q (2y 2 - 0) dy r y dy = I [y ] 18 *- x 20. f(y) = 4 - y 2 , g(y) = 0, c = -2, d = 2 =* A = f [f(y) - g(y)] dy = £(4 - y 2 ) dy 4y 2(8 32 x=4-y 2 21. Let us find the intersection points: ^- = %±- => y 2 - y - 2 = => (y- 2)(y + 1) = => y or y ■1, d = 2; f(y) = y-f , g(y) A = f [f(y) - g(y)] dy = £ (*±* - £ ) dy f £ l (y + 2-y 2 )dy=i[£+2y-f '"' H(i 4 L I 2 + 4 3 ; ^2 z ^ 3 I)] 22. Let us find the intersection points: ^— = y ^ => y 2 - y - 20 = => (y - 5)(y + 4) = => y = -4 y + 16 y 2 -4 ory = 5 =* c = -4, d = 5; f(y) = ^ , g(y) - A =* A = £ [f(y) - g(y)] dy = J^^ 16 - ^) dy 5£(y + 2 °-y 2 ) d y = 3p + 20y-f I[(f + 100-f)-(f-80+f)] 4 1 f9 4 (1 + 180-63) 1 /9 117) = i(9 + 234) 243 x= (y+16)/4 23. f(x) = x, g(x) = sin x, a = 0, b = | n b p 7r/4 I [f(x) — g(x)] dx = I (x — sin x) dx t./ >i ' Jo I 7I "/ 4 / 2 A + «,]r-(s + ^) i y=x y=sinx -* — x jt/4 - b = - 2' u 2 24. f(x)= l,g(x)= |sinx|,a =>• A = f [f(x) - g(x)] dx = f^ (1 - |sin x|) dx i/ a t/ — 7r/2 (1 + sin x) dx + I (1 — sin x) dx = 2 / (1 — sin x) dx = 2[x + cos x] ' 2(1-0 7T-2 Copyright (c) 2006 Pearson Education 1 — X 348 Chapter 5 Integration 25. a = 0, b = 7r, f(x) — g(x) = 2 sin x — sin 2x =>- A = I (2 sin x — sin 2x) dx = [—2 cos : = [_2.(-l)+i]-(-2-l + I)=4 cos 2x 1 ^ 26. a |, b = |, f(x) — g(x) = 8 cos x — sec 2 x c - (-8 • ^ + a/3) = 6x/3 pT/3 ^g A = / (8 cos x — sec 2 x) dx = [8 sin x — tan x.] 71 ' '' */ — 7T/3 tt/3 2 y=8cosx =(secx) 2 i x 27. f(y) = ^y, g(y) = 2-y,c = l,d = 2 =► A = / [f(y) - g(y)] dy = J] [^/y - (2 - y)] dy = / 2 (^-2 + y)dy= [| y 3 / 2 - 2y + | V 2 - 4 + 2 ;i-2+i; 7 _ 8\/2-7 6 6 28. f(y) = 6 - y, g(y) = y 2 , c = 1, d = 2 => A = j [f(y) - g(y)] dy = [ey-*-*]'=(i2-2-| = 4 / 2 (6 - y - y 2 ) dy 7 , 1 _ 24-14+3 3^2 6 13 6 1- y=VL r=6-x -i — X 4 5 29. f(x) = x 3 - 3x 2 = x 2 (x - 3) =>• f (x) = 3x 2 - 6x = 3x(x - 2) =$■ f = +++ | | ++- 2 =>• f(0) = is a maximum and f(2) = —4 is a minimum. A = — / (x 3 — 3x 2 ) dx = — $ — j 27 4 30. A (f-27) JJ(a 1/2 - XV2) 2 dx = £{a - 2 v ^x 1 /2 + x ) dx = [ax - \ y/i: f (6 - 8 + 3) = f 3/2 I jr 2, 1 1 3^2; — a — j -y/a • a^a 3 1 . The area above the x-axis is Ai = I (y 2//3 — y) dy jq ; the area below the x-axis is £<> 2/3 _ y) dy 3y°/' 3 5 11 10 the total area is Ai + A 2 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Chapter 5 Practice Exercises 349 X7r/4 n 57r/4 (cos x — sin x) dx + I (sin x — cos x) dx J 5tt/4 (cos x — sin x) dx = [sin x + cos x] [- cos x - sin x] ,' 4 + [sin x + cos x] 57r y 4 7T/4 y = sin i = [(f + #) -(o+i)] + [{4 + 4)-{-4-4)} + [(-1+0)- (-4-4)} = *~4 -2 = 4^2-2 33. y = x 2 + f*\ dt ={► | = 2x + 1 => g = 2 - + ; y(l) = 1 + £ \ dt = 1 and y'(l) = 2+1=3 34. y = J (1 + 2^/sect) dt =>• & = 1 + 2^/sec x =>• = 2 Q) (sec x) _1 / 2 (sec x tan x) = ^/sec x (tan x); (l + 2^/sec t) dt = and x = =4> 35. y= J^dt-3 => g = ^;x = 5 => y = //** 36. y = J \/2 - sin 2 1 dt + 2 so that ^ = y/l - sin 2 x; x = - 1 => y = J \f % = l + 2v^0 = 3 dt - 3 = -3 2 - sin 2 t dt + 2 = 2 37. Let u = cos x =>• du = —sin x dx =>• — du = sin x dx J 2(cos x)-V2 S i n x dx = J 2iT 1 / 2 (- du) = -2 J iT 1 / 2 du = -2 (j?£) + C = -4U 1 / 2 + C = -4(cos x) 1 / 2 + C 38. Let u = tan x =>• du = sec - x dx J (tan x)- 3 / 2 sec 2 x dx = J u" 3 / 2 du = f^ + C = -2U- 1 / 2 + C = j^]- 2 + C 39. Let u = 26> + 1 => du = 2 d0 =5> § du = d0 J [20 + 1 + 2 cos (20 + 1)] d0 = J (u + 2 cos u) (± du) = f + sin u + Ci - , = 2 + + sin (20 + 1) + C, where C = Ci + J is still an arbitrary constant 2=t^-Lsin(20+ 1) + Ci 40. Let u = 20 - 7T => du = 2 d0 => ± du = d0 /(y3b + 2sec2 ^- 7r >) d0 = /(7n + 2sec2u )(5 du ) = 5/K 1/2 + 2 = \ (^-\ + \ (2 tan u) + C = u 1 / 2 + tan u + C = (20 - tt) 1 / 2 + tan (20 - tt) + C sec" u) du J< 41. I (t-|) (t-Hf)dt: J(t 2 -|)dt=/(t 2 -4r 2 )dt=f-4(^)+C=f + t+C 42. J^^dt=/^dt=J, P + ?)* : /o r 2 + 2r 3 ) dt : (-i) (S) +C=-i- i - i + C t f2 + *" 43. Let u = 2t 3 / 2 => du = 3^ldt =>• |du = i/t dt J ^/tsin (2t 3 / 2 )dt = 1 J sin u du = -jcos u + C = -±cos(2t 3 / 2 ) + C Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 350 Chapter 5 Integration 44. Let u = 1 + sec 6 => du = sec 9 tan# 69 =>■ J sec 6 tmO \/l + sec 9 69 = J u^du = §u 3 / 2 + C §(l + sec6>) 3/2 +C 45. 46. 47. J (3x 2 - 4x + 7) dx = [x 3 - 2x 2 + 7x] ^ = [l 3 - 2(1) 2 + 7(1)] - [(-l) 3 - 2(-l) 2 + 7(-l)] = 6 - (-10) = 16 J q (8s 3 - 12s 2 + 5) ds = [2s 4 - 4s 3 + 5s] J = [2(1) 4 - 4(1) 3 + 5(1)] -0 = 3 /V 4 / 3 dx = [-3X- 1 / 3 ] J 7 = -3(27)^/3 - (-3d)- 1 / 3 ) = -3 (§) + 3(1) = 2 r*=r^=r r>/8 *=[- 2ri/s ]j -2 _ (-2) 50. Let x = 1 + Ju => dx = \ ir 1 / 2 du => 2dx=-^=;u=l => x = 2, u = 4 => x = 3 J' ii± ^ ! du = £ xV2( 2 dx) = [2 (|) x 3 / 2 ] J = 4 (3 3 / 2 ) - I (2 3 / 2 ) = 4^3 - | v^ = f (3 v/3 - 2^2 51. Letu = 2x+ 1 =>• du = 2 dx =>• 18 du = 36 dx; x = => u= l,x = 1 => u = 3 r^ = r ^ du = w ! = ^ i = (^ - (t*) 52. Let u = 7 - 5r =>■ du = -5 dr =4> - i du = dr; r = =^> u = 7, r = 1 =>• u = 2 £ 1/ ^ W = £ (7-5rr 2 / 3 dr=/ 7 2 u- 2 / 3 (-Idu) = -I[3u 1 / 3 ] 2 7 =|( 3 v ^- 3 v ^ 53. Let u = 1 - x 2 / 3 => du = - | x" 1 / 3 dx => - | du = x" 1 / 3 dx; x = | =>• u = 1 - (|) 2/3 = | 1 => u = 1 - l 2 / 3 = £ x " /3 ( J - x2/3 ) 3/2 dx = C u3/2 ( " i du) = [ ( - ^ (f )1 L = [" i u5/2 ] 1/8 27^ 160 3/4 3 ( )5/2 !) G) 5/2 54. Let u = 1 + 9x 4 => du = 36x 3 dx => ^ du = x 3 dx; x = => u = 1, x = ± => u = 1 + 9 Q) 4 J o 1/2 X 3 (1+9xr 3/ 2dx= f /16 U -3/ 2 (l du ) = [1 (^)]; 5/16 = r X _-!/»! -A- = _ J, (25\-l/2 _(_±. (l)-l/2) = J_ 18 U6^ V 18 KL > I 90 n 4 _ 25 16 J_ -1/2] 25 18 u J 1 55. Let u = 5r => du = 5 dr =4> j du = dr; r = =>■ u = 0, r = 7r => u = 57r £sin 2 5rdr= J"(sin 2 u) (§ du) = § [§ - ^] ^ = (| - ^) - (0 - ^) = f 56. Let u = 4t - | =4> du = 4 dt =>• j du = dt; t = =^ u 3tt £t=-=>U— - 4 ' l 4 ^ u 4 p?r/4 /»37r/4 / cos 2 (4t-|)dt= I (cos 2 u)(idu) = i[ t/ <-/ — 7T/4 i [u , sin2u ] 37r /4 _ i /^ sin(f) \ _ 1 / tt , sin (- |) \ 4 L2 " r 4 J -tt/4 ~~ 4 ^ 8 T 4 J 4 \ 8' 4 ^ 7T J_ I J_ ZI 8 16 ' 16 ~~ 8 Copfiigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Chapter 5 Practice Exercises 35 1 nw/3 57. J sec 2 6d9 = [tan 8] n < 3 = tan f - tan = \fl 58. J»37r/4 C rr/4 esc 2 x dx = [-cot xfj' = (- cot &) - (- cot |) = 2 59. Let u = | =>- du = ^ dx =>• 6 du = dx; x = 7r => u = g, x = 3n =>- u = | DO o 2 J cot 2 fdx=J 6cot 2 udu = 6j (csc 2 u - 1) du = [6(-cot u - u)]^g = 6 (- cot f - f) - 6 (-cot | - f) 6^3 -2tt 60. Let u = | =^ du = ± d(9 =>■ 3 du = &6; 9 = =>• u = 0, 6» = tt => u = f J tan 2 fd6>=J (sec 2 f - l) dff = J 3 (sec 2 u - 1) du = [3 tan u - 3u]£ = [3 tan f - 3 (f )] - (3 tan - 0) = 3a/3 - tt tt/3 61. I sec x tan x dx = [sec x]^ , 3 = sec — sec (— |) = 1 — 2 = — 1 -?r/3 >3tt/4 J»3 csc z cot z dz = [—esc z]J){ = f— esc 3 ' ^ 7T/4 J 7t/4 csc I) = -V2- '2 = 63. Let u = sin x => du = cos x dx; x = => u = 0, x = f =$■ u=l ■>jr/ 2 J 5(sin x) 3 / 2 cos x dx = J 5u 3 / 2 du = [5 (|) u 5 / 2 ] Q = [2u 5 / 2 ] Q = 2(1) 5 / 2 - 2(0) 5 / 2 = 2 64. Let u = 1 - x 2 => du = -2x dx => - du = 2x dx; x = - 1 => u = 0, x = 1 => u = i: 2x sin (1 — x ) dx I' — sin u du = 65. Let u = sin 3x => du = 3 cos 3x dx =>• | du = cos 3x dx; x f_ 15sin 4 3xcos3xdx= J 15u 4 (± du) = J 5u 4 du = [u 5 ] 1 : u = sin i &) = l,x=f => u = sin(f) (-1) 5 -(D 5 66. Letu = cos (|) =4> du = - \ sin (|) dx => -2 du = sin (§) dx; x = =4> a = cos (§) = 1, x = ^f =3- u = cos ( -§-J l 2 "•27T/3 // cos- 4 (|)sin(f)dx=j; "u- 4 (-2du)=[-2(^)]| 2 =| ( I)- 3 -|(i r 3 = 2 (8 _ 1)= i4 67. Let u = 1 + 3 sin 2 x =>• du = 6 sin x cos x dx =>■ | du = 3 sin x cos x dx; x = =>• u=l,x=| => u = 1 + 3 sin 2 |=4 X V2 7ms dx = f * (i du ) = r s u " /2 du = [Kf)l ! = t ul/2 ] I = 4l/2 - il/2 = l 68. Let u = 1 + 7 tan x =>■ du = 7 sec 2 x dx =>■ i du = sec 2 x dx; x = =>■ u = 1 + 7 tan = 1, x = | => u = 1 + 7 tan f = 8 /r W f^*=r*(H«)=ri"- 2/s < i »=[Kf)]!=[? ui/! ']i=' <8) ' /3 -' (i > i/s =' C§io)t (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 352 Chapter 5 Integration 69. Let u = sec 9 => du = sec 9 tan 9 d$: 9 = =$ u = sec = 1, u = sec f=2 /.ir/3 Jo tan i/ 2 sec i d0 Jo 5 2 LC-i) sec 9 v 2 sec 2 d9 = r^^k dS = l 1 du- -4- -s« 3/2 du /2u J 1 n/2(2) I x/2Tl)J '2-1 70. Let u = sin ^t => du = (cos 0) (± r 1 / 2 ) dt = ^ dt => 2 du = ^ dt; t = ^ 6 2 t = ^ => u = sin § = 1 J ir 2 /36 -^^ dt =j r i 2 7u( 2du )= 2 J r l u " /2du =[ 4 v / ^]: /2 = 4 ^ i - 4 yi= 2 ( 2 -^ yt sin x/t 71. (a) av(f) = i^T) /_', (mx + b)dx = \ [s£ + bx] ^ = 1 [(sQ£+b(l)) - (=*=£+ b(-l))] = i (2b) = b (b) av(f) = ^ £ (mx + b) dx = £ [=£ + bx] ^ = £ [("f 2 + b(k)) - (=*=* + b(-k))] = £ (2bk) = b 72. (a) y av = ^/^dx = \f ^^' 2 dx = f [| x 3 / 2 ] „ = f [f (3) 3 / 2 - § (O) 3 / 2 ] = f (2^) = 2 (b) y« = ^0 T v^ dx = s/>* 1/2 dx = 4 [1 x3/2 ] o = ^(! ^ 3/2 - I (°) 3/2 ) = £ (I a v^) = 1 a J ^/axf '(x) dx = g^ [f(x)]J = ^ [f(b) - f(a)] = ^ff^ so the average value of f ' over [a, b] is the slope of the secant line joining the points (a, f(a)) and (b, f(b)), which is the average rate of change of f over [a, b]. 73. f' = JT- av b — a 74. Yes, because the average value of f on [a, b] is g-^ I f(x) dx. If the length of the interval is 2, then b — a = 2 and the average value of the function is | I f(x) dx. 75. We want to evaluate •*365 305 p365 />365 / s^oJo f W dx =355j 37sin l^-io 1 ) 25 dx = J£ 305 r Jo sin 2- 505 (x- 101) dx 25 305 I' Jo dx Notice that the period of y = sin |^ (x — 101 2^ 575 is -r = 365 and that we are integrating this function over an iterval of length 365. Thus the value of 305 Jo 2- :in.-. (x-101) n365 dx + H I dx is J£ ■ + H ■ 365 = 25. 365 J 365 365 76 - 675^20 /r^ 8 " 27 + 10_5 ( 26T - L87T2 )) dT = 6^5 8.27T - 26T 2 _ 1.87T 3 2-10 5 3-10 5 675 1 655 .27(675) 26(675f 1.87(675)-' 2-10 5 3-K^ 07On\ -1- 26 ( 20 ) 2 _ 1 - 87 (2Q) 3 ■ Z 'W 1" 2-105 3 . 10 5 20 ■ J « ^(3724.44 -165.40) = 5.43 = the average value of C v on [20, 675]. To find the temperature T at which C v = 5.43, solve 5.43 = 8.27 + 10~ 5 (26T - 1.87T 2 ) for T. We obtain 1.87T 2 - 26T - 284000 = 26±,/(26) 2 -4(1.87)(-284000) _ 2 6±y / 212 4996 2(1.87) interval [20, 675], so T = 396.72°C. 3.74 -. So T = -382.82 or T = 396.72. Only T = 396.72 lies in the J> V^ 77. g = \/ 'i + cos-'x Copffigl (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle Chapter 5 Practice Exercises 353 78. g = v/2 + cos 3 (7x 2 ) • |^(7x 2 ) = 14x^2 + cos 3 (7x 2 ) 7Q dy _ _d_ ly - dx — dx Ji 3 + t dt 6 3+x 4 80. | A fr 2 i dx \ J sec x t2 + 1 dt /; p+i dt 1 d / scc \ _ sec x tan x sec 2 x + 1 dx *■ * 1 + sec 2 x 81. Yes. The function f, being differentiable on [a, b], is then continuous on [a, b]. The Fundamental Theorem of Calculus says that every continuous function on [a, b] is the derivative of a function on [a, b]. 82. The second part of the Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) on nb [a, b], then I f(x) dx = F(b) — F(a). In particular, if F(x) is an antiderivaitve of y 1 + x 4 on [0, 1], then J q y4 +x 4 dx = F(l) - F(0). 83. y 84. y r 1 + 1 2 dt = - 1 + 1 2 dt dy _ _d_ dx dx /.x 1 + t 2 dt d_ dx /iX 1 + 1 2 dt -yTT^ dt r r^ dt dy _ dx dx r pcosx r pcosx E [~ J„ T^t2 dt = ~ dx" |y o ~ dt . 1 — cos 2 x 7 V dx E ( cos x >) t\-) (— sin x) = -J— = esc x m^ x / v y sin x 85. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of the parking lot on each interval by averaging the widths at top and bottom. This gives the estimate Ass 15 • ^ + 36 , 36 + 54 , 54 + 51 , 51 + 49.5 , 49.5 + 54 , 54 + 64.4 , 64.4 + 67.5 , 67.5 + 42 ^2" l "2~ t "2" t "2" r 2" ,_ 2 _r 2 '2 >2 XKo ™ot io A^ao . f"<tO 1fl/ft2\ r- /"<QA1 ff2\ f<£1 in/fi-2\ ') A«5961ft 2 .ThecostisArea-($2.10/ft 2 )« (5961 ft 2 ) ($2.10/ft 2 ) = $12,518.10 => the job cannot be done for $11,000. 86. (a) Before the chute opens for A, a = —32 ft/sec 2 . Since the helicopter is hovering, Vo = ft/sec => v = J -32 dt = -32t + v = -32t. Then s = 6400 ft => s = J -32t dt = -16t 2 + s = -16t 2 + 6400. At t = 4 sec, s = -16(4) 2 + 6400 = 6144 ft when A's chute opens; (b) For B, s = 7000 ft, v = 0, a = -32 ft/sec 2 =4> v = J -32 dt = -32t + v = -32t =4> s = J -32t dt = -16t 2 + s = -16t 2 + 7000. At t = 13 sec, s = -16(13) 2 + 7000 = 4296 ft when B's chute opens; (c) After the chutes open, v = - 16 ft/sec => s = J - 16 dt = - 16t + s . For A, s = 6144 ft and for B, s = 4296 ft. Therefore, for A, s = - 16t + 6144 and for B, s = - 16t + 4296. When they hit the ground, for A, 0= -16t + 6144 => t 6144 16 384 seconds, and for B, = - 16t + 4296 => t 4296 16 268.5 seconds to hit the ground after the chutes open. Since B's chute opens 58 seconds after A's opens > B hits the ground first. p30 87. av(I) = i J (1200 - 40t) dt = i [1200t 20t 2]30 Jo — 30 ^ [((1200(30) - 20(30) 2 ) - (1200(0) - 20(0) 2 )] ±5 (18,000) = 600; Average Daily Holding Cost = (600)($0.03) = $18 3(1 /»14 av(I) = n J (600 + 600t) dt = ^ [600t + 300t 2 ]J = ± [600(14) + 300(14) 2 - 0] = 4800; Average Daily Holding Cost = (4800)($0.04) = $192 Cop # (c| 1 Pen Educate k, publishing as Pearson Addison-Wesle 354 Chapter 5 Integration 89. av(I) = jof () (450 - I) dt = i f450t - f 1 = i [450(30) - ^ - o] = 300; Average Daily Holding Cost = (300)($0.02) = $6 90. av(I) = i J o ^600 - 20-/l5t) dt = gg £ (600 - 20\/l5 l 1 / 2 ) dt = i [600t - 20^15 (§) t 3 / 2 = ^ [600(60) - ^^ (6O) 3 / 2 - Ol = i (36,000 - (2|Q) 15 2 ) = 200; Average Daily Holding Cost = (200)($0.005) = $1.00 CHAPTER 5 ADDITIONAL AND ADVANCED EXERCISES 1. (a) Yes, because f f(x) dx = ± f 7f(x) dx = ± (7) = 1 (b) No. For example, J o 8x dx = [4x 2 ] J = 4, but J] V8x dx = [2^ (^l)l = ^ (l 3 / 2 - : 3/2) 1 dx = — 3 2. (a) True: J f(x) dx = - J f(x) 1 (b) True: fjfQQ + g(x)] dx = /_,f(x) dx + J^gW dx = J_/(x) dx + J] f(x) dx + J_ ? g(x) dx =4+3+2=9 (c) False: J f(x) dx = 4 + 3 = 7 > 2 = J g(x) dx => J [f(x) - g(x)] dx > => J*_ [g(x) - f(x)] dx < 0. On the other hand, f(x) < g(x) =>■ [g(x) — f(x)] > =£• I [g(x) — f(x)] dx > which is a contradiction. 3. y = i f Q f(t) sin a(x - t) dt = lf g f(t) = ssls J"f(t) cos at dt - £21M J" f(t) sin ax cos at dt — dx cos ax sin at dt cos at dt sin ax / _d_ a I dx sin at dt d^v dx-' J f(t) cos at dt J + sin ax J^ f(t) sin at dt - seis* | A J f( t ) , cos ax J f(t) cos at dt + ^f^ (f(x) cos ax) + sin ax J f(t) sin at dt - ^^ (f(x) sin ax) > ^ = cos ax I f(t) cos at dt + sin ax I f(t) sin at dt. Next, = —a sin ax I f(t) cos at dt + (cos ax) I ~ I f(t) cos at dt ) + a cos ax I f(t) sin at dt (sin ax) J>' J> sin at dt = — a sin ax / f(t) cos at dt + (cos ax)f(x) cos ax + a cos ax I f(t) sin at dt + (sin ax)f(x) sin ax = —a sin ax I f(t) cos at dt + a cos ax I f(t) sin at dt + f(x). •J Jo Jo Therefore, y" + a 2 y = a cos ax I f(t) sin at dt — a sin ax I f(t) cos at dt + f(x) ^ J o f(t) cos at dt - £2|S J f(t) sin at dt J = f(x). Note also that y'(0) = y(0) = 0. 4. x = SI 7^ * =* = (X) = ^IoV^ dt =r y [£ 7^ dt] (I) from the chain rule |(1 + 4yT 1/2 (8y) (g) = ^M = M^l = 4 y. Thus g = 4y, and the constant of Copy |t (c| 1 Pearson Etati, Inc., publishing as Pearson Addison-Wesle Chapter 5 Additional and Advanced Exercises 355 proportionality is 4. 5. (a) £ f(t) =► f(x 2 ) (b) I t 2 dt «7 dt = X COS 7TX =>• ^ I f(t) dt = COS 7TX — 7TX sin 7TX =^> f (X 2 ) (2x) = COS 7TX — 7TX sill 7TX COS 7TX — 7TX Sin 7TX Tl-1.10 v O — ^ fY/1 \ COS 2tT — 27T SIB 2?T 1 2x Thus, x = 2 ) 3 - '?, i, ,. _ 3 f(x) , I =5 (f(x)) 3 ■5 I 3 «4) - \ (f(x)) 3 = X COS 7TX => (f(x)) 3 = 3x COS 7TX f(x) = VJ X COS 7TX => f(4) = V 3(4) cos 4tt = \A2 6. J f(x) dx = | + | sin a + § cos a. Let F(a) = J f(t) dt => f(a) = F'(a). Now F(a) = f + § sin a + § cos a | + | sin | =>■ f(a) = F (a) = a + ± sin a + | cos a - | sin a =4> f (| 7. J] f(x) dx = x/b 2 + l - ^2 =► f(b) = & J^ f(x) dx = 1 (b 2 + l) _1/2 (2b) W- cos 5 - I sin s - - 1 _ w _ 1 2 2 ""* 2 2 ' 2 2 2 x/b^TT f(x) vA^+T 8. The derivative of the left side of the equation is: -jj- I I f(t) dt du = / f(t) dt; the derivative of the right side of the equation is: £■ I f(u)(x — u) du = ^ I f(u) x du — -jj- I u f(u) du H [*J> )du _d_ dx J u u f(u) du = J f(u) du + x \£ J f(u) du - xf(x) = j^ f(u) du + xf(x) - xf(x) = I f(u) du. Since each side has the same derivative, they differ by a constant, and since both sides equal J when x = 0, the constant must be 0. Therefore, I I f(t) dt du = I f(u)(x — u) du. 9. !| = 3x 2 + 2 => y = J (3x 2 + 2) dx = x 3 + 2x + C. Then (1,-1) on the curve => l 3 + 2(1) +C = -1 =4> C = — : => y = x 3 + 2x - 4 10. The acceleration due to gravity downward is —32 ft/sec 2 =4- v = I —32 dt = — 32t + Vo, where Vo is the initial velocity =>• v = -32t + 32 => s = J (-32t + 32) dt = -16t 2 + 32t + C. If the release point, at t = 0, is s = 0, then C = => s = -16t 2 + 32t. Then s = 17 => 17 = -16t 2 + 32t => 16t 2 - 32t + 17 = 0. The discriminant of this quadratic equation is —64 which says there is no real time when s = 17 ft. You had better duck. n3 p0 ni 11. I f(x)dx= / x 2 / 3 dx+ I -4dx J -8 J -8 -4x] 3 [l^ 3 ]^ (0 - | (-8) 5 / 3 ) + (-4(3) - 0) 36 5 f -12 dx y = x 2/1 4 2 -8 -4 3 X -4' L y = -4 — • 12. £f(x) dx = £^ dx + £ (x 2 - 4) = [-f(-^) 3/2 ]- 4 +[f-H[ (-l(4) 3 / 2 )] + [(f-4(3))-0] Lo-(- 16 _ o _ Z 3 J — 3 Copffigl (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle 356 Chapter 5 Integration ■ /. g(t) dt = •J t dt + 1 sin 7rt = [«:+ [-1 1 2 COS 7TtJ = (J-o) + [- - - cos 2ir -(- 1 2 2 TT i dt 14. J h(z) dz = J v 7 ! -z dz + J (7z - 6)" 1 / 3 dz = [-§(!- z)3/ 2 ] J + [^ (7z - 6)^3] J = [-|d- D 3/2 -(-Id- O) 3 / 2 )] "£ (7(2) - 6) 2 / 3 - i (7(1) - 6) 2 / 3 ] 1 L 14 2 , /"6 6 _ i_) — 55 7 14/ — 42 15. J f(x) dx = J '_ dx + J (1 - x 2 ) dx + J = w:£ + [x - f ] ] i + [2x] 2 (-l-(-2)) + (l-f)-(-l-^ 2 ) + [2(2) -2(1)] 1 + l)+4-2=f 16. J h(r) dr = J r dr + J (1 - r 2 ) dr + J dr ,, + wi (0-^) + ((l-f)-0) + (2-l) -1+1+1=1 y=l y=2 v-1 17. Ave. value = j^/Vd) dx = ji, £f(x) dx = 1 [/Jx dx + Jj'(x - 1) dx] = 1 [*] V 1 [$ - x] ' = »[(¥-o) + (*-2)-(f-i)] = j 18. Ave. value = ^ J] f(x) dx = 3+0 £ f(x) dx = | |£ dx + Jj dx + £ dx = i [1-0 + + 3- 2] 19. f «=£Td^fw=Hi)-(i)aa)) = ^-x(-^) = ^+Hi 2a f W = £°Jr^ dt =* f '«=(HiK)(s( sinx ))-(i^^)(s( cosx )) cos x sin x cos x 1 sin x cos 2 x sin 2 x 21. g(y) = / v . y sint 2 dt =► g'(y)= (sin(2 v ^) 2 ) (| ( 2> /y)) - (sin(^) 2 ) (|(^y)) sin 4y sin y 22. f(x) = f + t(5 - t) dt => f (x) = (x + 3)(5 - (x + 3)) {i (x + 3)) - x(5 - x) (g) = (x + 3)(2 - x) - x(5 - x) = 6 - x - x 2 - 5x + x 2 = 6 - 6x. Thus f (x) = =4> 6-6x = =>• x=l. Also, f"(x) = -6 < =4> x = 1 gives a Copffigl (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Chapter 5 Additional and Advanced Exercises 357 maximum. 23. Let f(x) = x 5 on [0, 11. Partition [0, 1] into n subintervals with Ax = — = -. Then 1, -,...,- are the v J l j j l j j n n n ' n ' ' n oo / . \ 5 right-hand endpoints of the subintervals. Since f is increasing on [0, 1], U = J2 \i) (n) * s me u PP er sum f° r f(x) = x 5 on [0, 1] => lim f] ( ±-Y (±) = lim ± [(i) - J'" 1 Jo CD : lim n — > oo l° + 2^ + ... + ir p6 5 dx - . . o L 6 J o 24. Let f(x) = x 3 on [0, 1]. Partition [0, 1] into n subintervals with Ax 1-0 n n i Then ±, = , ... , i are the oo / ■ \ 3 right-hand endpoints of the subintervals. Since f is increasing on [0, 1], U = J2 [ i ) („) * s me u PP er sum f° r f(x) = x 3 on [0,1] => lim oo / . \ 3 E (i) (J) lim i (1} n — > cc n \\ n ) (2) 3 '| :_ Jim 1^ n - r'x 3 Jo dx 25. Let y = f(x) on [0, 1]. Partition [0, 1] into n subintervals with Ax = — - = -. Then -, -,...,- are the - 7vyL,J l:j nn n'n''n right-hand endpoints of the subintervals. Since f is continuous on [0, 1], £) f ( - ) Q) is a Riemann sum of y = f(x)on[0,l] => n lim^Ef(0(D= n lim o \ [f (±) + f (f) + •■ • + f (=)] = £ f(x) dx 26. (a) Jjm^ ^[2 + 4 + 6+... + 2n] = n Em, 1 [2 + 4 + 5 + .. . + f] = J^x dx = [x 2 ]J = 1, where f(x) = 2x on [0, 1] (see Exercise 25) <W n^oo ^[l 15 + 2 15 + -+» 15 ] = n !H n oc ^[a) 15 + (D" + -+ (S) "]= X> ^ = [^] ^ = ^, where f(x) = x 15 on [0, 1] (see Exercise 25) (c) lim - [sin - + sin — + ...+ sin — 1 = I sin mi dx = [— - cos 7rxl . = — - cos 7r — (— - cos 0) = -, where f(x) = sin n\ on [0, 1] (see Exercise 25) (d) lim 4, [1 15 + 2 15 + ... +n 15 l = ( lim A ( lim -^ \l 15 + 2 15 + . . . + n 15 }) = ( lim ±) f x 15 dx = (yg) =0 (see part (b) above) (e) lim 4i [1 15 + 2 15 + ... +n 15 l = lim -& fl 15 + 2 15 + .. . + n 15 l v ' n — > oo n lj L J n _> oo n 16 L J = (^iPoo n ) U^F 1 Sie I 1 " + 215 + ■■■ +n 15 ]) = LliPlo n ) J x 15 dx = oo (see part (b) above) 27. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal to r, the radius of the circle) and a vertex angle of 8„ where 9 a = — . The area of each triangle is A n = i r 2 sin 6 n =4> the area of the polygon is A = nA„ = ^- sin 6„ t&in^. (b) lim A = lim ™f s i n 2* = li m 2^ sin 27T _ i; m („ r 2\ si "(v) _ ,..,.:', ,;,„ v„ : r i ^ lim (m 2 ) ^^ = (ttt 2 ) lim 27r/n -> 28. y = sinx+ I cos2tdt + 1 = sinx — I cos 2t dt + 1 =4> y' = cos x — cos(2x); when x = tt we have y' = cos7r — cos(27r) = — 1 — 1 = — 2. And y" = — sinx + 2sin(2x); when x = 7r, y = sin7r + I cos2tdt + 1 + 0+1 = 1. Cfiojt (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 358 Chapter 5 Integration dt = 29. (a) g(l) = I f(t (b) g(3) = / 3 f(t)dt=-i(2)(l) = -l (C) g(-l) = / 'f(t)dt=-j\(t)dt: dt = 0=>x >2 2 ) = -tt -3, 1, 3 and the sign chart for g'(x) = f(x) is -+ | | +++. So g has a 1 3 dt (d) g'(x)=f(x) relative maximum at x = 1 . (e) g'(-l) =f(-l) =2 is the slope and g(-l) = J f(t) y = 2x + 2 - 7T. (f) g"(x)=f'(x)=0atx inflection point for g at x = — 1. We notice that g"(x) = f'(x) < for x on (— 1, 2) and g"(x) = f'(x) > for x on (2, 4), even though g"(2) does not exist, g has a tangent line at x = 2, so there is an inflection point at x = 2. (g) g is continuous on [—3, 4] and so it attains its absolute maximum and minimum values on this interval. We saw in (d) that g'(x) = => x = -3, 1, 3. We have that -71", by (c). Thus the equation isy + 7r = 2(x+l) -1 and g"(x) = f'(x) is negative on (—3, —1) and positive on (— 1, 1) so there is an g(-3) = f~ 'fCt g(l) = J>( g( 4 ) = Ij( dt= - l ) dt = ,-(t)dt=-^ 2 -2tt t) dt ■1 t dt: ■1-1 Thus, the absolute minimum is — 2ir and the absolute maximum is 0. Thus, the range is [— 2ir, 0]. Copyright (c) 2006 Pearson Education Chapter 5 Additional and Advanced Exercises 359 NOTES: Copyright (c) 2006 Pearson Education, 360 Chapter 5 Integration NOTES: Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS 6.1 VOLUMES BY SLICING AND ROTATION ABOUT AN AXIS 1. (a) A = 7r(radius) 2 and radius = y 1 - x 2 =>- A(x) = tt (1 — x 2 ) (b) A = width • height, width = height = 2\/l - x 2 =>■ A(x) =4(1 - x 2 ) (c) A = (side) 2 and diagonal = y^side) =>- A = (diag ° nal)2 ; diagonal = 2\/l - x 2 =>• A(x) = 2 (1 - x 2 (d) A = ^ (side) 2 and side = 2\/l - x 2 => A(x) = a/3 (1 - x 2 ) 2. (a) A = 7r(radius) 2 and radius = -y/x => A(x) = 7rx (b) A = width - height, width = height = 2^/x => A(x) = 4x (c) A = (side) 2 and diagonal = y^Cside) =4> A = (diag ° nal)2 ; diagonal = 2^/x =>■ A(x) = 2x (d) A = ^ (side) 2 and side = 2\/x =>• A(x) = a/^x 3. A(x) = (dlas ° nal)2 = (^-(-\A)) 2 = 2x (see Exercise lc); a = 0, b = 4; V = f A(x) dx = J 2x dx = [x 2 ] * = 16 4. A (x) = " (dla 7 ter)2 = " 1(2 - x p- x2]2 = " [2(1 - x2) ' 2 = tt (1 - 2x 2 + x 4 ) ; a = -1, b = 1; V = J] A(x) dx = J_ t tt (1 - 2x 2 + x 4 ) dx = tt I" x-|x3 + ^^=2.(l-| + i) = if 5. A(x) = (edge) 2 = \yf\ - x 2 - (-\/l - x 2 )] = (ly/i - x 2 V =4(1 - x 2 ) ; a = -l,b= 1; V = £ A(x) dx = f_ t 4(1 - x 2 ) dx = 4 |x - f 1 =8(1- i) 16 3 6. A(x) (diagonal)- al) 2 _ [fi^-{-vv^)\ 2 _ (is/r^y 2(1— x 2 ) (see Exercise lc); a = — 1, b = 1; V = J A(x) dx = 2J_ t (1 - x 2 ) dx = 2 [ *■ T|_ 1 =4(l-I) = | 7. (a) STEP 1) A(x) = \ (side) • (side) - (sin |) = \ ■ (l\Jsmx\ ■ [l\/sm x\ (sin |) = \fb sin x STEP 2) a = 0, b = tt STEP 3) V = J a A(x) dx = v/3 JJsin x dx = V-y/l cos xj " = \fl{\ + 1) = 2^/3 (b) STEP 1) A(x) = (side) 2 = (ly/sin x\ (ly/sm x\ = 4 sin x STEP 2) a = 0, b = tt STEP 3) V = J A(x) dx = f 4 sin x dx = [-4 cos x] * = 8 8. (a) STEP1) A(x) = ^^£_£ = | ( S ec x - tan x) 2 = f (sec 2 x + tan 2 x - 2 sec x tan x) [sec 2 x+(sec 2 x-l)-2^] 4 STEP 2) a=-f,b 3' " ~~ 3 b rirft STEP 3) V = J a A(x) dx = f | (2 sec 2 x - 1 - ^f ) dx = f [2 tan x - x + 2 (- ^ -)V /3 x/J-tt/3 Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 362 Chapter 6 Applications of Definite Integrals = l[2 V ^-! + 2(- I } y )-(-2y3 + f+ 2(- I j y ))]=|(4y3^) (b) STEP 1) A(x) = (edge) 2 = (sec x - tan x) 2 = (2 sec 2 x - 1 - 2 |j£) STEP 2) a= -f,b= f STEP 3) V = £A(x) dx = £^ (2 sec 2 x - 1 - f$i) dx = 2 (2^3 - f ) = 4 ^3 - f 9. A(y) = | (diameter) 2 = | (\/5y 2 - o) = & y 4 ; c = 0, d = 2; V = / c A(y) dy = £ 5 f y 4 dy = [(f) (f ) E M5 (2 5 - 0) = 8tt 1.5 0.5 o+ — j — r~ s 10. A(y) = i (leg)(leg) = 1 [^l^f - (-/T^y 2 )] 2 = \ {l^T^ff = 2 (1 -y 2 ) ; c = -1, d = 1; V = / c d A(y)dy = /_ 1 i 2(l-y 2 )dy = 2[y-^^=4(l-I) = | 11. (a) It follows from Cavalieri's Principle that the volume of a column is the same as the volume of a right prism with a square base of side length s and altitude h. Thus, STEP 1) A(x) = (side length) 2 = s 2 ; STEP 2) a = 0, b = h; STEP 3) V = J A(x) dx = J q s 2 dx = s 2 h (b) From Cavalieri's Principle we conclude that the volume of the column is the same as the volume of the prism described above, regardless of the number of turns =>• V = s 2 h 12. 1) The solid and the cone have the same altitude of 12. 2) The cross sections of the solid are disks of diameter x— (|) = |. If we place the vertex of the cone at the origin of the coordinate system and make its axis of symmetry coincide with the x-axis then the cone's cross sections will be circular disks of diameter | — (— |) = | (see accompanying figure). 3) The solid and the cone have equal altitudes and identical parallel cross sections. From Cavalieri's Principle we conclude that the solid and the cone have the same volume. 13. R(x) =y=l-|=>V= XV[R(x)] 2 dx = n£(l - |) 2 dx = tt/^I - x + f ) dx = tt [x - f + £ 7T(2- 2 ' UJ 3 14. R(y) = x = | =► V = £ ^[R(y)] 2 dy = ^£{^f dy = tt £\ y 2 dy = tt [| ,31 I = n . 3 . g = ^ 15. R(x) = tan (| y) ; u = | y => du = | dy => 4 du = tt dy; y = => u = 0, y = 1 => u = f ; V = f o 7r[R(y)] 2 dy = n£ [tan (f y)] 2 dy = 4 £ "tan 2 u du = 4 £ (-1 + sec 2 u) du = 4[-u + tan u]q /4 Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 6.1 Volumes by Slicing and Rotation About an Axis 363 4(-f + l-0)=4-7r C T / 2 r, 16. R(x) = sin x cos x; R(x) = => a = and b = | are the limits of integration; V = I 7r[R(x)] dx Trf* (sin x cos x) 2 dx = tt J* (si " 4 2x)2 dx; [u = 2x => du = 2 dx =4> | = ^;X = => U = 0, u = 7 r] - V = ^/;isin 2 udu=|[f-isin2u]; = f [(f-0)-0] 17. R(x) = x 2 =4> V = f~ ?r[R(x)] 2 dx = tt / ~(x 2 ) 2 dx 7 r/Vdx = 7 rff 32tt 5 18. R(x) = x 3 =4> V = J tt[R(x)] 2 dx = wj (x 3 ) 2 dx 1287T 7 19. R(x) = ^9-x 2 =4> V = J_ 3 7r[R(x)] 2 dx = tt f (9 - x 2 ) dx tt 9x 2tt [9(3) - f ] = 2 • tt • 18 = 36tt 20. R(x) = x - x 2 =>• V = J o tt[R(x)] 2 dx = ttJ o (x - x 2 ) 2 dx = tt£{v? - 2x 3 + x 4 ) dx = TT U 2x1 J_ J! 3 4 + 5 .3 2 ' 5^ 30 ^(10- 15 + 6)= ^ 30 '■tt/ 2 21. R(x) = ^/cosx =>• V = J* 7r[R(x)] 2 dx = tt J* cos x dx tt [sin x] q = 7r(l — 0) = 7T y=(cosx) 1/2 Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 364 Chapter 6 Applications of Definite Integrals 22. /tt/4 /»7t/4 7r[R(x)] 2 dx = 7t / sec 2 x -7r/4 J —tt/4 dx 7r[tanx]!^ /4 = 7r[l-(-l)]=27r 1 0.8 0.6 23. R(x) = \fl - sec x tan x => V = £ tt[R(x)] 2 dx fV 4 / i- \ 2 = 7r / v2 — sec x tan x dx Jo \ v / = 7r I ( 2 — 2 v 2 sec x tan x + sec 2 x tan 2 x J dx ( r n/4 r p 7 ^ 4 n /4 ■? ■> \ = 7r I / 2dx — 2\/2j sec x tan x dx + I (tan x) sec x dx I <>•'• \ / 0.2 = n f[2x]^ /4 - 2 V / 2[sec x]f + [^] ^ = ^[(|-0)-2y2(y2-l)+i(l 3 -0)]=7r(f + 2y2-^) y = V2 y = sec i tan x 24. R(x) = 2 - 2 sin x = 2(1 - sin x) =>■ V = F ~tt[R(x)] 2 dx ** X7r/2 f*V 2 4(1 — sin x) 2 dx = 47r I (1 + sin 2 x — 2 sin x) dx = 47rJ" r [l + 1 (1 - cos 2x) - 2 sin x] dx ^L ( 4tt 4 x - | _ co|2x _ 2 sin x j i2x + 2 cos x tt/2 L2 "■ 4 ' - — ""JO = 4tt [(f - + 0) - (0 - + 2)] = ttOtt - 8) 25. R(y) = v/5 • y 2 =* V = £ ^[R(y)] 2 dy = it £ 5y 4 dy = 7T [y 5 ] ^ = tt[1 - (-1)] = 2tt 1 x *■ X 26. R(y) = y 3 / 2 => V = J^ ^[R(y)] 2 dy = ^y 3 dy 47T 27. R(y) = V2 sin 2y =* V = £ ^[R(y)] 2 dy r /2 r n - = 7r / 2 sin 2y dy = ir [— cos 2yJ j = tt[1 -(-1)] = 2tt i;' 2 Copyright (c) 2006 Pearson Education Section 6.1 Volumes by Slicing and Rotation About an Axis 28. R(y) = ^/cosf => V = J%[R(y)] 2 dy y = 7T /° 2 COS (f ) dy = 4 [sin f ] °_ 2 = 4[0 - (-1)] = 4 365 29. R(y) = ^ =* V = |>[R(y)] 2 dy = 4* £ ^ dy 47T y+i 47r[-i -(-!)] =37 * = 2/(»+l) 30. R(y) = ^ => V = J o ' 7r[R(y)] 2 dy = ir£ 2y (y 2 + 1)~ 2 dy; [u = y 2 + 1 =^> du = 2y dy; y = =4> u = 1, y = 1 => u = 2] - v = tt £ u- 2 du = tt [- 1] ; = n [- 1 - (-D] = i x = V^/(s/ , + l) TJ3 5!3 53 BT" 31. For the sketch given, a = - § , b = § ; R(x) = 1, r(x) = v /cosx; V = J] tt ([R(x)] 2 - [r(x)] 2 ) dx = 1 7r(l — cos x) dx = 27r I (1 — cos x) dx = 27r[x — sin x]q = 27r (| — l) = 7r 2 — 2tt 32. For the sketch given, c = 0, d = f ; R(y) = 1, r(y) = tan y; V = J tt ([R(y)] 2 - [r(y)] 2 ) dy = tt£'\i - tan 2 y) dy = tt £'\l - sec 2 y) dy = 7r[2y - tan y]^ 4 =7r(f-l) = ^-7T 33. r(x) = x and R(x) = 1 =>- V = / 7T ([R(x)] 2 - [r(x)] 2 ) dx XV(l-x 2 )dx = .[x-f] o =^[(l-I)-0] 277 3 34. r(x) = 2^/x and R(x) = 2 =>■ V = J 7r ([R(x)] 2 - [r(x)] 2 ) dx = 7 r/ o 1 (4-4x)dx = 4 7 r[x-f]^4 7 r(l-i)=27r Copy fight (c| 1 Pearson Etati, he, publishing as Pearson Addison-Wesle 366 Chapter 6 Applications of Definite Integrals 35. r(x) = x 2 + 1 and R(x) = x + 3 => V = J%([R(x)] 2 -[r(x)] 2 )dx = 7r£ [(x + 3) 2 -(x 2 + l) 2 ]dx = 7T f [(x 2 + 6x + 9) - (x 4 + 2x 2 + 1)] dx = 7T f (-x 4 - x 2 + 6x + 8) dx i x ., _ £ _ £ + 6x! + g x 7r [(-f-| + M + 16) _ ( i + I + |_ 8)]=7r( _ f _ 3 + 28-3 _ 1 5-3 0-33 \ Wir 36. r(x) = 2 - x and R(x) = 4 - x 2 => V = /_", ^([R(x)] 2 -[r(x)] 2 )dx = *£ [( 4 - x2 ) 2 - ( 2 - x > 2 ] dx = 7T /_* [(16 - 8x 2 + x 4 ) - (4 - 4x+ x 2 )] dx = n J (12 + 4x - 9x 2 + x 4 ) dx 7T 1 12x + 2x 2 - 3x 3 + f = 7T [(24 + 8 - 24+ f ) - (-12 + 2 + 3 - i)] = 7T (15 37. r(x) = sec x and R(x) = y 2 (2 — sec 2 x) dx = 7r[2x — tan x]_ , 4 = 7r[(f-l)-(-f + l)]=7r(7r-2) 33 \ _ 108tt y y = secx Ht/4 */4 38. R(x) = sec x and r(x) = tan x => V = J%([R(x)] 2 -[r(x)] 2 )dx = 7r I (sec 2 x — tan 2 x) dx = 7r I 1 dx = 7r[x]J = 7r 39. r (y) = 1 and R(y) = 1 + y => V = J%([R(y)] 2 -[r(y)] 2 )dy = ir£ [(1 + y) 2 - 1] dy = 7T £ (1 + 2y + y 2 - 1) dy = .X 1 (2y + y 2 )dy = 7 r[y 2 + ^;=.(l + I) = 4 y =x-1 < X Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle Section 6.1 Volumes by Slicing and Rotation About an Axis 367 40. R(y) = 1 and r(y) = 1 - y =>• V = f tt ([R(y)] 2 - [r(y)] 2 ) dy *J = irf o [1 - (1 - y) 2 ] dy = tt £ [1 - (1 - 2y + y 2 )] dy = n I ( 2 y ~ y 2 ) d y = ^ [> \y 2 - y i TT 1 n _ 2tt 41. R(y) = 2andr(y) = ^y => V = f Q 7T ([R(y)] 2 - [r(y)] 2 ) dy = ttJ q (4 - y) dy = tt Uy - y] = tt(16 - 8) = 8tt 42. R(y) = v/3 and r(y) = ^3 - y 2 =* V = / o 3 tt ([R(y)] 2 - [r(y)] 2 ) dy = it jf [3 - (3 - y 2 )] dy = tt ^ y 2 dy ;rl T ttV^ t^x 43. R(y) = 2 and r(y) = 1 + ^/y => V = / o 1 7r([R(y)] 2 -[r(y)] 2 )dy = -r[ 4 -0 + ^) 2 ] d y = 7r / o 1 (4-l-2^/y-y)dy = ^X'( 3 - 2 ^-y) d y 4 3/2 _ r 2 Jo M 6 J = 7?r " 6 tt |3y - f y - (3 - f " | 44. R(y) = 2 - y 1 / 3 and r(y) = 1 => V = J%([R(y)] 2 -[r(y)] 2 )dy = -X l [( 2 -y 1/3 ) 2 - 1 ] d y = t:£ (4 - 4y 1 / 3 + y 2 / 3 - l) dy = tt f (3 - 4y 1 < /3 + y 2 / 3 ) dy = tt [3y - 3y 4 / 3 + ^1 * = tt (3 - 3 + 3^r 5 c y =-x3 Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 368 Chapter 6 Applications of Definite Integrals 45. (a) r(x) = ^/x and R(x) = 2 =>■ V = / ) 4 7r([R(x)] 2 -[r(x)] 2 )dx = 7T J u (4 - x) dx = 7T Ux - f 1 = 7T(16 - 8) = 8tt (b) r(y) = and R(y) = y 2 => V = £7r([R(y)] 2 -[r(y)] 2 )dy r 2 = 7T J o y 4 dy = 7T 32g 5 (c) r(x) = and R(x) = 2 - yfi. => V = £ it ([R(x)] 2 - [r(x)] 2 ) dx = tt£ (2 - ^/x") 2 dx 7rJ (4-4-y/x + x) dx = 7r 4x- 8x 3 / 2 i x 2 3 " l " 2 7T(16 64 _, 16^1 8tt 3 "r" 2 ) ~ 3 1 X (d) r(y) = 4 - y 2 and R(y) = 4 => V = £ tt ([R(y)] 2 - [r(y)] 2 ) dy = tt J^ [l6 - (4 - y- f | dy 2 tt / o (16 - 16 + 8y 2 - y 4 ) dy = it J q (8y 2 - y 4 ) dy = tt [f y 3 _ r 5 '64 32 \ _ 224?r 46. (a) r(y) = and R(y) = 1 - § =*• v = X, 2 * ( [R( y )]2 - [r( y )]2 ) d y = -X 2 ( 1 -D 2d y = -r( 1 -y + T)dy (b) r(y) = 1 and R(y) = 2 - \ - V = £ tt ([R(y)] 2 - [r(y)] 2 ) dy = tt £ [(2 - \f - l] dy = tt £ (a - 2y + J - l) dy -r(3-2y+£)dy = 7 r[3y-y 2 + 1 C=.(6-4+A) =7r (2 + !) 8tt 37 — 3 47. (a) r(x) = and R(x) = 1 - x 2 =* V = /_'^([R(x)] 2 -[r(x)] 2 )dx = tt J_* (1 - x 2 ) 2 dx = tt £ (1 - 2x 2 + x 4 ) dx = 7r[x-f + ^] ^ = 27T (1 - | + I) 2,(15^3) = m 1 (b) r(x) = 1 and R(x) = 2 - x 2 =J> V = £ tt ([R(x)] 2 - [r(x)] 2 ) dx = tt £ \(2 - x 2 ) 2 - lj dx = tt £ (4 - 4x 2 + x 4 - 1) dx = tt£ (3 - 4x 2 + x 4 ) dx = tt hx - f x 3 + f 1 = 2tt (3 - f + ±) ff (45 - 20 + 3) = Sgr (c) r(x) = 1 + x 2 and R(x) = 2 =>■ V = J*_ [ tt ([R(x)] 2 - [r(x)] 2 ) dx = tt £ U - (1+ x 2 ) 2 ] dx = tt /_' (4 - 1 - 2x 2 - x 4 ) dx = tt£ (3 - 2x 2 - x 4 ) dx = tt hx - § x 3 - f 1 = 2tt (3 - 2 _ I N 3 5/ £(45-10-3)= Sfe Copffigl (c| 1 Pearson Educate he, publishing as Pearson Addison-Wesle Section 6.1 Volumes by Slicing and Rotation About an Axis 369 48. (a) r(x) = and R(x) = - j) x + h => V = £7r([R(x)] 2 -[r(x)] 2 )dx =^r(-s x+h ) 2dx *£( f 2 x 2 - ^ x + h 2 ) <lx ;.-lr 1^- y+x|" - -..lr f 1 : b + b) ^h 2 b 3 =-(h/b)x+h (b) r(y) = OandR(y) = b(l-£) =>■ V = ]% ([R(y)] 2 - [r(y)] 2 ) dy = Trb 2 /^ (l - I) 2 dy 49. R(y) = b + ,/a 2 - y 2 and r(y) = b - ^/a 2 - y 2 =>■ V=/_ a a 7r([R(y)] 2 -[r(y)] 2 )dy = * /_", [( b + V^f) 2 - (b - y^^7) 2 ] dy = 7T J_ 4b^/a 2 - y 2 dy = 4b7rJ_ i/a 2 - y 2 dy = 4b7r • area of semicircle of radius a = 4b7r • ^f- = 2a 2 b7r 2 50. (a) A cross section has radius r = ^/2y and area 7rr 2 = 27ry. The volume is I 27rydy = tt [y 2 ] = 257T. l dy A(h) ' dt ' (b) V(h) = /A(h)dh, so f = A(h). Therefore f = f • f = A(h) ■ f , so f - ! dV For h = 4, the area is 2tt(4) = 8?r, so f = gL • 3 i^ _3_ units 3 dt 87r " sec 8tt sec 51. (a) R(y) = v'a 2 - y 2 => V = tt J " (a 2 - y 2 ) dy = tt |"a 2 y - = tt [a 2 h - i (h 3 - 3h 2 a + 3ha 2 - a 3 ) - f 1 = tt (a 2 h - f + h 1 ^| a 2 h _ a 3_(h^a£_ (_ a 3 01 a — ha' -h 2 (3a-h) 3 (b) Given ^ = 0.2 m 3 /sec and a = 5 m, find § | h=4 . From part (a), V(h) =* ^ = 107rh-rf => ^ = dv.^=^h(10-h)^ =* dt ~~ dh dt 7rh 2 (15-h) 3 0.2 57Th 1 2 ?rh 3 dh I dt I h=4 4tt(10-4) (20tt)(6) 120tt k m/sec. 52. Suppose the solid is produced by revolving y = 2 — x about the y-axis. Cast a shadow of the solid on a plane parallel to the xy-plane. Use an approximation such as the Trapezoid Rule, to P 2 estimate / 7r[R(y)] dy « J^7r k-l Ay. 53. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius h has been removed. Thus its area is Ai = 7rR 2 — 7rh 2 = tt (R 2 — h 2 ) . The cross section of the hemisphere is a disk of radius \/R 2 — h 2 . Therefore its area is A-2 = tt ( y R 2 — h 2 j = tt (R 2 — h 2 ) . We can see that Ai = A2. The altitudes of both solids are R. Applying Cavalieri's Principle we find Volume of Hemisphere = (Volume of Cylinder) - (Volume of Cone) = (ttR 2 ) R - \tt (R 2 ) R = § ttR 3 . Copyright (c| 1 Pearson E Won, Inc., publishing as Pearson Addison-Wesle 370 Chapter 6 Applications of Definite Integrals 54. R(x) = £ => V = XV[R(x)] 2 dx = tt/^ dx = f [l h2 ) ( T ) = 5 7rI ' 2 h, the volume of a cone of radius r and height h. 55. R(y) = V256 - y 2 =► V = / ^[RCy)] 2 dy = tt/ J256 - y 2 ) dy = n [256y - £ = 7T [(256)(-7) + f - ((256)(-16) + 1 f)] = * (y + 256(16 - 7) - ^) = 1053tt cm 3 « 3308 cm 3 56. R(x) = fr 2 \/36 - x 2 => V = £ tt[R(x)] 2 dx = irf* £ (36 - x 2 ) dx = fo £ (36x 2 - x 4 ) dx 12 12x 3 144 ^ 12 - 6 3 _ £ 5 y 144 ^ 12 f) 196tt \ / 60-36 1 367i cm 3 . The plumb bob will weigh about W = (8.5)(^p) « 192 gm, to the nearest gram. 57. (a) R(x) = |c — sin x| , so V = 7r I [R(x)] 2 dx = it I (c — sin x) 2 dx = tt I (c 2 — 2c sin x + sin 2 x) dx tt/* (c 2 - 2c sin x + 1 ~ c ° s2x ) dx = tt£ (c 2 + | - 2c sin x - cos 2x\ dx = tt [(c 2 + \) x + 2c cos x - Siii] = tt [(c 2 tt + | - 2c - 0) - (0 + 2c - 0)] = tt (c 2 tt + | - 4c) . Let V(c) = n (c 2 7T + I — 4c) . We find the extreme values of V(c): ^ = 7r(2c7r — 4) = => c = -^ is a critical point, and V (f ) = tt (£ + f - f ) = tt (f - £) = ^ - 4; Evaluate V at the endpoints: V(0) = y and V(l) = 7r (| 7r — 4) = \ — (4 — 7r)7r. Now we see that the function's absolute minimum value is ^ — 4, taken on at the critical point c = -. (See also the accompanying graph.) 2 (b) From the discussion in part (a) we conclude that the function's absolute maximum value is y , taken on at the endpoint c = 0. (c) The graph of the solid's volume as a function of c for < c < 1 is given at the right. As c moves away from [0, 1] the volume of the solid increases without bound. If we approximate the solid as a set of solid disks, we can see that the radius of a typical disk increases without bounds as c moves away from [0, 1]. 58. (a) R(x) = 1 - f| =► V = /_ 4 4 7r[R(x)] 2 dx = -[*-i + 5^] 4 _ 4 = 2 -( 4 -! + 5^) 2tt(4 ! + !) 24 ' 5-1 (60 - 40 + 12) 64?r ft 3 15 ll 1-(x2/16) (b) The helicopter will be able to fly (^f ) (7.481)(2) « 201 additional miles 6.2 VOLUME BY CYLINDRICAL SHELLS 1. For the sketch given, a = 0, b = 2; V = J> (X s ) (hth.) d X = />X (l + *) dx = 27T/; (X + £) dx = 27T [i + i] ' = 27T = 27T • 3 = 67T 2 T lb) 2. For the sketch given, a = 0, b = 2; V = J> (** ) (*J) dx = />rx (2 - f ) dx = 27r/; ( 2 x - f ) dx = 2tt [x 2 £ 16 2tt(4 - 1) = 6tt Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 6.2 Volume by Cylindrical Shells 371 3. For the sketch given, c = 0, d = y2; v = r^(^)(^ t ) d y = X V "^y(y 2 )dy = 2^y 3 dy = 2.p]f = 2. 4. For the sketch given, c = 0, d = y 3; •v^> v^ v = X 2. ( ** ) (St) d y = /. 2 -y ■ p - ( 3 - y 2 )] d y = 2 - X ' y 3 d y = 2 - [J 5. For the sketch given, a = 0, b = y 3; v = J> (X.) (it) d * = J^™ • (v 7 ^) dx; | u = x 2 + 1 =4> du = 2x dx; x = =4> u = 1, x = \/3 => u = 4 1 -> V = ^ u 1/2 du = 7T [| u 3 / 2 ] J = f (4 3 / 2 - 1) = (f ) (8 - 1) 41 v/3 9^ 2 14- 3 6. For the sketch given, a = 0, b = 3; — J a Vradius/ ^ height^ ~~ J \^/y? + 9 ) ' [u = x 3 + 9 =>• du = 3x 2 dx => 3 du = 9x 2 dx; x = : -^ V = 2tt /"3U- 1 / 2 du = 6. [2u J / 2 ] ^ 6 = 12tt (y^6 J,x = 3 = 36tt u = 36] 7. a = 0, b = 2; V = J> (** ) (**) dx = X 2 2.x [x -(-!)] dx = X, 2 2ttx 2 - 1 dx = 7T X 3x 2 dx = 7T [x 3 ] \ = 8tt a = 0, b= 1; V = j> GEL) (S) ^ = X, 1 2 ™ ( 2x " 1) dx = .X' 2 (f)dx = .i3x 2 dx = 7 r[x 3 ]; = . 9. a = 0,b= 1; V = X> (££) (£t) dx = X' 2 ™ K2 - x) - x 2 ] dx = 2ttX (2x - x 2 - x 3 ) dx = 2tt [x 2 - f - 2^(1-1-^ 2tt ' 12-4-3'" IOtt _ Stt 12 — 6 y=2-x Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 372 Chapter 6 Applications of Definite Integrals 10. a = 0,b= 1 V = J> ( ** ) (**) dx = / o 2.x [(2 - x*) - x 2 ] dx = 2tt £ x (2 - 2x 2 ) dx = 4tt/ o (x - x 3 ) dx = 47T Ml-i) 11. a = 0,b= 1; V = j> (££) (S) ^ = /.' 2 ™ [^ - (2X - 1} ] ^ = 2ttJ* (x 3 / 2 - 2x 2 + x) dx = 2tt [§ x 5 / 2 3 X + 2 X J Q 2ir 3 ' 2 J 2tt ' 12-20+ 15 1 30 ; 7rr 15 12. a= l,b = 4; V = J> ( r S) (it) dx = J> x (I x " /2 ) dx = 3tt J^x 1 / 2 dx = 3ir [f x 3 / 2 ] \ = 2ir (4 3 / 2 - l) = 2tt(8 - 1) = 14tt 2 1.5 + 1 0.5 tf5 !TS 1 2Vx" 13. (a) xf(x) = <^ => xf(x) = <^ ^ x, x = ^ sin x, < x < 7r 0, x = since sin = we have xf(x) = | sin x, < x < 7r sin x, x = =>- xf(x) = sin x, < x < 7r (b) V = f\w (*£) ( *gj dx = Io 2lTX ■ f(x) dx and x • f(x) = sin x, < x < 7T by part (a) =>• V = 2ir J sin x dx = 2ix\— cos x]J = 27r(— cos 7r + cos 0) = 47r 14. (a) xg(x) <x<| x = ,2 v n < x < tt/4 x = xg(x) { tan 2 x, < x < 7i74 0, x = since tan = we have 1 x-0, x f tan 2 x, xg(x) = i . 2 [ tan~ x, x (b) V = fl-K (*£) (*St) dx = JJ^ttx • g(x) dx and x • g(x) = tan 2 x, < x < tt/4 by part (a) => V = 2tt J 7 " tan 2 x dx = 2?r J (sec 2 x - 1) dx = 2?r[tan x - x][/ 4 = 2tt (l xg(x) = tan 2 x, < x < 7r/4 •>tt/4 At; — 7t 2 Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 15. c = 0,d = 2; v = J> (SJ (S) d y = J>y [^ - (-y)] d y =2tt r Jo y 2 ) dy = 2?r v y . j y ^ - - s 5 ^(3^ + 5) 2tt (?# + f) = 16.(f + i) Section 6.2 Volume by Cylindrical Shells y 373 2 x=-y ^ljp #x => /y -2 vk 16. c = 0, d = 2; v = j> GEL) (S) ^ = T 2 -y [y 2 - (-y)]dy 2^(y 3 + y 2 ) d y = 2 ^fT+T 167r(| + i) \6tt 1 4()tt t— X 17. c = 0, d = 2; v = J> ( r S) (St) d y = 1 2 -y ( 2 y - y 2 ) d y = 2^;(2y 2 - y 3)dy = 2 7 r[¥-fl' = 2 -(f^) 32tt « 12 3 x=2y-y : 1 x 18. c = 0, d= 1; v = J> (£.) (S) d y = /.' 2 -y ( 2 y - y 2 - y) d y = 2tt£ y (y - y 2 ) dy = 2n£ (y 2 - y 3 ) dy 2n\\-\ ^{\-\) 19. c = 0,d= 1; v = J> (SJ (S) d y = 2 -X' y[y - (-yw = 2^ , 2y 2 dy=f [ y 3]J 47T 10 3 20. c = 0, d = 2; v = r 2 -( r ^)(S) d y = r 2 -y(y-D d y = 2 -r^y=f[y 3 ] 1 _ 8tt — 3 Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 374 Chapter 6 Applications of Definite Integrals 21. c = 0, d ml V = I 2- ( ** ) (**) dy = / o 2.y [(2 + y) - y 2 ] dy = 2n £ (2y + y 2 - y 3 ) dy = 2. [y 2 + £ - 2tt 4 16 ±2 = | (48 + 32 - 48) 16- 3 22. c = 0, d = 1 V = I 2n (** ) (**) dy = X 27 ry [(2 - y) - y 2 ] dy = 2n£ (2y - y 2 - y 3 ) dy = 2. [y 2 - f - Mi-l-i) (12-4-3) 6 23. (a) V = X 2. ( shell \ / shell radius / \ height dy = X' 2ny - 12 (y 2 - y 3 ) dy = 24. £ (y 3 - y 4 ) dy = 24. [£ - f ] * 24. ' i r 24_7T _ 67T 20 — 5 w v = X> (Xs) (h s S0 d y = 1 2 ^ - y) t 12 (y 2 - y 3 )] d y = 24 -X' o - y) (y 2 - y 3 ) d y = 24.X 1 (y 2 -2y 3 + y 4 )dy = 24 7 rp-^ + ^; = 24.(1-1 + I) =24.(3^) = f (o v = X> (It) (fit) d y = X' 2 - (! - y) I 12 (y 2 - y 3 )i d y = 24 - /.' (I - y) (y 2 - y 3 ) d y 24 ^X 1 (sy 2 -Ty 3 + y 4 ) d y = 247r [ 8 „3 13 „4 20 y 4 + 24.! 13 , n 20 T 5 y 2-1;; 60 (32 - 39 + 12) 2. (d) V = ./>(**) ( h ^ t ) dy = X' 2^ (y + I) [12 (y 2 - y 3 )] dy = 24n£ (y + §) (y 2 - y 3 ) dy = 24 -X (y 3 - y 4 + I y 2 - I y 3 ] d y = 24 ^X (I y 2 + 1 y 3 - y 4 ) d y = 24. | ^ >■■■ + :n >< ■ 2 „3 1 3 „4 2^(^ + h 5) = W (8 + 9 - 12) = 2 t = 2. 24. (a) V = X 2* (*£ ) ( -J) dy = X 2.y f? - (J - £)] dy = J~ 2.y (y 2 - $) dy = 2^ (y 3 - £) dy = 2 7 r[^-g] 2 o =2 7 r($-|) = 32* (J - £) = 32* (J - $) = 32* (£) = f (b) V = X> (*£) (** ) dy = X>(2 - y) [£ - ($ - £ )] dy = £ 2.(2 - y) (y 2 - £) dy = 2 -X 2 ( 2 y 2 -^-y 3 + fldy = 2-[¥-^-^ + fl^ = 2-(f-i-f + |) = f (0 v = J> (,££) (5t) d y = 1 2 -( 5 - y) [? - (S - ?)] dy = X^ 5 - y) (y 2 - 1) d y = 2 -r(V~!y 4 -y 3 + ?)dy = 2 7 r[f-|f-^ + ^] 2 = 2.(f-if-if + |)=8 7 r (d) v = X^( r r)(he; g L)dy = X 22 -(y + ep-p-fl]dy = X 22 -(y + I)(y 2 -^)dy 2tt f ( v 3 - y - + - v 2 - - v 4 ^l dv = 2. \ y - - ^ + ^f - Ml = 2. (16 _ w + 40 _ 160\ = 4 Jo V 4 ^ 8 y 32 y J u > ^" [4 24 T 24 160j „ ^ U 24 ^ 24 160^ ^" Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Section 6.2 Volume by Cylindrical Shells 375 25. (a) About x-axis: V= J>(*£) (*j)dy = I" 2 ^y(yy - y)dy = ^/J (y 3/2 - y 2 ) d y = 2-[§y 5/2 -|y 3 ]: = 2.(!-i) = i Abouty-axis:V=/>( l S s )(h5t) dx = J 2ttx(x - x 2 )dx = 2?r J (x 2 - x 3 )dx 2tt i = 27r (3-l) (b) About x-axis: R(x) = x and r(x) = x 2 => V = J tt[R(x) 2 - r(x) 2 ]dx = J tt[x 2 - x 4 ]dx .3 5J 2£ 1") About y-axis: R(y) = ,/y and r(y) = y => V = J -k [R(y) — r(y) ] dy = J 7r[y — y 2 ]dy | f _ yf " ' 2 3 ,2 3^ 26. (a) V = J\r[R(x) 2 -r(x) 2 ]dx = 7rJ ) 4 [(f + 2) 2 -x 2 ]dx = TrJ^-fx 2 + 2x + 4)dx = 7r[-f + x 2 + 4x1 = tt(-16 + 16 + 16) = 16tt (b) V = /> ( ** ) (**)«* = />rx(§ + 2 - x)dx = lW-i)dx = 2^ o 4 (2x-f)dx < X (c) V = />(*£) (ht".) dx = SM A " x )(f + 2 - *)d* = J>(4 -x)(2 - |)dx = 2^(8 - 4x + % )dx 2tt 8x - 2x 2 27r(32-32 + 64 \ _ 64?r (d) V = _{ i b 7r[R(x) 2 -r(x) 2 ]dx = 7r/ o 4 [(8-x) 2 - (6- | ) 2 ]dx = 7r^ 4 [(64 - 16x + x 2 ) - (36-6x+^) irf Q (|x 2 -10x + 28)dx = 7r[^ - 5x 2 + 28x1 =tt[16- (5) (16) + (7) (16)] = tt(3)(16) = 48tt dx 27. (a) V = J> (** ) ( -J) dy = J>y(y - I) dy = 2^ 2 (y 2 - y) dy = 2, [f - f ] ' 2tt (I - 2 + i) = | (14 - 12 + 3) = % 2 (b) V = J%(**)(**)dx = f*2-irx(2 - x) dx = 2ir£(2x - x 2 ) dx = 2tt [x 2 - f 1 = 2tt [(4 - |) - (l - ±)] = 2.[(^)-( 3 T l)]=2 7 rG-f) = f w v = /> (.s) (it) dx = r 2 - (t - x ) ( 2 - x ) dx = 2 -r '20 _ 16 . 3 3 x + x 2 ) dx 2.[fx-fx 2 + Ix3] 2 = 2.[(f-f + f)-(f-| + I)]=2.(|)=2. (d) V=/ c d 27T( shell \ / shell radius / \ height dy = f*2v(y - l)(y - 1) dy = 2^/^y - l) 2 = 2tt [ (y-D J 3 Copffigl (c| 1 Pearson E tati, Inc., publishing as Pearson Addison-Wesle 376 Chapter 6 Applications of Definite Integrals 28. (a) V = J> ( ** ) (**) dy = filiytf - 0) dy = 2.J;y 3 dy = 2. [^ = 2.(^=8. (b) V=/>(^j(^ t )dx = f*2nx (2 - y/x) dx = 2?r/ o (2x - x 3 / 2 ) dx = 2.[x 2 -§x 5 / 2 ] 4 , = 2.(l6- 2 ^) 2.(16- f,- 5 2yr (80 - 64) = ^ (0 V = J> (5£) (5t) dx = I 2 ^ 4 " x ) ( 2 - V^) ^ = 2./> - 4xV 2 - 2x + x 3 / 2 ) dx 2lT [8x - | x 3 / 2 - x 2 + | x 5 / 2 ] J = 2. (32 64 16 + 64 \ 2^ §(240-320 + 192)= f (112) 224:. 15 (d) V=/ c d 27T( shell \ / shell radius / I height dy = ^2.(2 - y) (y 2 ) dy = 2. Jj2y 2 - y 3 ) dy = 2. [§ y 3 - £] 2tt 16 16\ 32. 12 (4-3) 8^ 3 29. (a) V = J> (** ) (** ) dy = J>y(y - y 3 ) dy = />(y 2 -y 4 )dy = 2.[^-^ = 2.(i-i) 4- 15 shell \ / shell (b) V = £2-( r X)^; g h t Jdy = / o 1 2.(l-y)(y-y 3 )dy x=y-y 3 2-X'(y-y 2 -y 3 + y 4 )dy = 2-p-^-? + fl; = 2^(i-i-i + i) = |(30-20- 15+12)= |g 30. (a) V = XM^)(^ t )dy = £ 2.y [1 - (y - y 3 )]dy = 2.j;'(y-y 2 + y 4 )dy = 2tt[£-£ + £ 2.(1-1+1) 117T 15 2^r 30 (15- 10 + 6) x=y-y 3 (b) Use the washer method: V = /V[R 2 (y)-r 2 (y)]dy = X 1 .[l 2 -(y-y 3 ) 2 ]dy = 7rj; ) 1 (l-y 2 -y G + 2y 4 )dy = .[ y -^-^ + f]^ . 1 ^(105-35- 15 + 42) 97- 105 (c) Use the washer method: V = £. [R 2 (y) - r 2 (y)] dy = £ . [[1 - (y - y 3 )] 2 - 0] dy = tt£ [l - 2 (y - y 3 ) + (y - y 3 ) 2 ] dy = SI ( = jfo (70 + 30 + 105 - 2 • 42) (d) V = J> (** ) (**) dy = £ 2.(1 - y) [1 - (y - y 3 )] dy = 2. £(l - y) (1 - y + y 3 ) dy (1 + y 2 + y G - 2y + 2y 3 - 2y 4 ) dy = .[y+f + ^-y 2 + f 1217T 210 .(l + I + I-l+I-§) 2 -J> - y + y 3 - y + y 2 - y 4 ! dy = 2ir£ (1 - 2y + y 2 + y 3 - y 4 ) dy = 2. v - ^ r 1 r _ r 3 ' 4 5 2.(1-1 + 1 + 1-1) |(20+15-12)- : ^ 30 Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle Section 6.2 Volume by Cylindrical Shells 377 31. (a) V = J> (** ) (**)dy = /; 2*y (^ - y») dy [4}/2 v 5/2 27r/ o (2v^y 3 /2 _ y a) dy = 27r [i*2 y ^ _ t 2tt M 2..4(f-l) _ ?1 8tt 5 27T (8 - 5) f 4-2 3 _ 44 \ V 5 4 ) 24- 5 W V = J> (*- ) ( -J) dx = /; 2.x (,/i - f ) dx = 2nj; (x 3 / 2 - f ) dx = 2n [§ x 4 5/2 3 _ tt-2 4 -3 _ 48;r 5 5 32. (a) V=J>(,X) (**)<* = J" 2ttx [(2x - x 2 ) - x] dx = 2tt J x (x - x 2 ) dx = 2?r J (x 2 - x 3 ) dx 2tt 2tt :*-« (b) V=/ a b 27T( shell \ / shell radius / I height dx = J 2tt(1 - x) [(2x - x 2 ) - x] dx = 2tt J (1 - x) (x - x 2 dx 2. J o (x - 2x 2 + x 3 ) dx = 2tt I 2 3 * + 4 2tt ' 2^ 2 3 ~ 4,/ § (6 - 8 + 3) 33. (a) V = f\ [R 2 (x) - r 2 (x)] dx = tt f 16 (x~ 1/2 - l) dx = vr [2xV2 _ x ] J /i6 = vr [(2 - 1) - (2 - I - ^)] 7T (1 6) 9^ 16 W V = X 2. (** ) ( -J) dy = J>y (£ - *) dy 1 v" 2 - Z- 2 ' 32 if (8 + i) 0.5 y = i 0^ 0:4 o.<s o:s — i - * 9tt 16 34. (a) V = f\ [R 2 (y) - r 2 (y)] dy = f\ ( ^ - i) dy = -[-|y- 3 -^i = -[(-^-l)-(-|-^)] = ^(-2-6+16 + 3)=^ w v = £2- (*-) (-J) dx = j>x (^ - 1) dx = 2 -/> i/2 - x ) dx =M§* 3/2 -C y 2- i- y-1/Vx .25 1/4 = 2 ^[(5-5)-(M-^)]=^(5- 1 -5+ii;) = 4|( 4 - 1 6-48-8 + 3) 35. (a) Disk: V = Vi - V 2 Vi = f V[Ri(x)] 2 dx and V 2 = f 2 7r[R 2 (x)] 2 with Ri(x) = A /^~ and R 2 (x) = Jx, J ai J ;i2 V ai = —2, bi = 1; a 2 = 0, b 2 = 1 => two integrals are required (b) Washer: V = V, - V 2 V! = J% ([Ri(x)] 2 - [ ri (x)] 2 ) dx with R x (x) = 11- 48 ^±2 and ri ( x ) = 0; ai = -2 and bi = 0; Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 32 378 Chapter 6 Applications of Definite Integrals V 2 = / ' 7T ([R 2 (x)] 2 - [r 2 (x)] 2 ) dx with R 2 (x) = y^ and r 2 (x) = y/l; a 2 = and b 2 = 1 => two integrals are required (c) Shell: V = £ 2n (** ) ( -J ) dy = £ 27 ry ( **) dy where shell height = f - (3y 2 - 2) = 2 - 2y 2 ; c = and d = 1. Only one integral is required. It is, therefore preferable to use the shell method. However, whichever method you use, you will get V = n. 36. (a) Disk: V = Vi - V 2 - V 3 Vj = J 7r[Ri(y)] 2 dy, i = 1, 2, 3 with Rj(y) = 1 and ci = -1, di = 1; R 2 (y) = ^/y and c 2 = and d 2 = 1; R3(y) = (~ y) 1 ^ 4 an d C3 = — 1, d3 = =>■ three integrals are required (b) Washer: V = Vi + V 2 V s = J\([R,(y)] 2 - [r(y)] 2 ) dy, i = 1, 2 with Ri(y) = 1, n(y) = y/y, Ci = and di = 1; R 2 (y) = 1> r 2 (y) = (— y) 1 ' 4 , c 2 = — 1 and d 2 = => two integrals are required (c) Shell: V=/>(^j(^ t )dx =/>x( h s e 1 ;: i : t )dx, where shell height = x 2 -(-x^) = x 2 + x 4 , a = and b = 1 => only one integral is required. It is, therefore preferable to use the shell method. However, whichever method you use, you will get V 6 • 6.3 LENGTHS OF PLANE CURVES 1. | = _ land £ = 3 Length = J^a/iOA = v/lO [t] 1-2/3 fVio 10 5/10 2. I = -sintand| = l+cost ^ J(f) 2 + (%)* = ^C" sin t) 2 + (1 + cos t) 2 = ^2 + 2 cos t => Length = / o V2 + 2cost dt = ^2 fiy/fc***) (1 + ^0 dt = v^ /J^^ dt = \/2 I — ^-SiSJ — dt (since sin t > on [0, 7r]); [u = 1 - cos t => du = sin t dt; t = => u = 0, v Jo s/\ -cost v — ' y t = 7T =>■ u = 2] -> a/2 /* u" 1 /' 2 du = a/2 [2U 1 / 2 ] * = 4 3. f =3t 2 and£=3t dx\ 2 , dtJ (I) = ^/(3t 2 ) 2 + (3t) 2 = ^9t 4 + 9t 2 = 3t Vt 2 + 1 (since t > on [o, v^] ) Length = J q St^/t 2 + 1 dt; [u = t 2 + 1 => § du = 3t dt; t = =>• u = 1, t = \/3 =>• u J i 4 |u 1 / 2 du= [u 3 / 2 ] 4 = (8 - 1) = 7 dx dy 4. f=tandf = (2t+l) 1 /2 => J(^) 2 4-(^) = Vt 2 + (2t 4- 1) = ^0+1)^ = |t 4- 1| = t + 1 since < t < 4 Length = J^(t + 1) dt = [f + t] = (8 4- 4) = 12 5. |=(2t + 3) 1 / 2 and^ = l+t =► ^ (f) 2 + {%)" = y/Qx + 3) + (1 + t) 2 = ^t 2 4- 4t 4- 4 = |t + 2| = t + 2 since < t < 3 => Length = J fl (t 4- 2) dt = [| + 2tl 2J_ 2 Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Section 6.3 Lengths of Plane Curves 379 6. f t = 8t cost and ^ = 8t sin t => J(f ) 2 + (^) = v/(8t cos t) 2 + (8t sin t) 2 = ^64t 2 cos 2 1 + 64t 2 sin 2 1 |8t| = 8t since < t < § =4> Length = f ~8t dt = [4t 2 ] " /2 - tt 2 dy _ I . 3 , v 2 , n1/2 dx 3 2 (x 2 + 2) V -2x= v/(x 2 +2)-: L = J v/l + (x 2 + 2)x 2 dx = JVl + 2x 2 +x 4 dx J o y(l+x 2 ) 2 dx=X(l+x 2 )dx=[x+f] o 3 + f = 12 -I — X dy _ 3 dx 2 L =X 4 V /r +^ dX; [ U=1 + 3 X du = | dx =>• | du = dx; x = => u=l;x = 4 => u = 10] -> L /■io , uV 2 du [| U 3/ 2 ]; 10 27 (lo^/To- l) Q ^ 2 7 - dy y 4y ,4 lit y 2 ^ (dy) 2 ' 16y 4 /,Vy 4 + 5 + i^ d y J,V(y 2 + ^)" d y = r(y 2 + ^) d y '27_I1_fl_lUQ_I_l i i-q i (-1-4 + 3) _ n , (-2) _ 53 3 12-/ V3 4/ 12 3 ' 4 io- | = |y 1/2 -b- 1/2 ^ (l) 2 = K y - 2 + rt =*L=j; ^l + i(y-2+i)dy = J,VK y+2+ O dy = -T* vW*)' d y = |/ 9 (y 1/2 + y" 1/2 ) dy = 5 [| y 3/2 + 2 y 1/2 ] J ,3/2 + y 1/2 ]; = (f + 3)-G + i) ii 1 _ 32 3 — 3 9- y#2 <■ 5 11. Jy - V 3_J_ =v fdx\ 2 6_1 , J_ — y 4y3 ^ \dy) - y 2 + 16y 6 >L =r v /i+y6 -5 + T^ dy J.y y° + 5 + T^ dy = /; y(y 3 + ^) 2 dy J/( y3 + ^) dy fie L_"\ _ (i-l) y_ _ y_ 4 8 1 1 , 1 _ 128-1-8+4 _ 123 32 4 32 32 Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 380 12. p dy Chapter 6 Applications of Definite Integrals = £-£ => (|) 2 = Hy 4 -2 + y - 4 ) > L = Xyi + |(y 4 -2 + y^)dy J 2 yi(y 4 + 2 + y-4)dy ^/;V(y 2 + y^ 2 ) 2d y = I IV + y~ 2 ) d y 1 r 2 3 I [(f - 1) - (! - 1)] = Hf - ! + 1) = 1 (6 + 1) 13 4 13. | =x l/3_I x -l/3 ^ ^ L =/ 1 V i + x2/3 m X 1 i X-2/3 2 ' 16 2/3 _ 1 2 t-2/3 l6~ dx dx J 1 V x2/3 +5+ /^(xVS-f l x -l/3) 2 dx = /'(xVS + 1 X-V3) dx r| x 4/3 + | x 2/3l8 3 r 2x 4/3. ,2/31 3 [( 2 . 2 4 + 2 2 ) - (2 + 1)] = | (32 + 4 - 3) = f 20" 15 y = 3x 4 ' 3 4 3x2,3 « 8 ^/ 10 5 1 H H»-X 14- t x 2 + 2x + 1 - (4x+4) 2 X 2 + 2x + 1 - i ! 4 (1+x) 2 (l+-) 2 -?(IW => (I) =d+*) 4 -2- + T6(lW l+x) 2 i (1+x)- dx; [u = 1 + x => du = dx; x = => u = 1, x = 2 =^ u = 3] 108-1-4+3 _ 106 _ 53 A) ~ " 12 12 6 15. dx dy V^ec 4 y-1 => (|) 2 = sec 4 y-l > L = J^ /4 v/l + (sec 4 y-l) dy = J^sec 2 y dy [tany]^ /4 =!-(-!) = 2 ..x= I Vsec 4 t -1 dt -91 /4- ->— x 16. dy Vi> m 3x 4 -l => L = f_' v/l + (3x 4 - 1) dx = /_,' a/3 x 2 dx '3 If v/3 l-(-2) 3 i/3(_l + 8) = ^ -2 y=JV3t 4 -1dt -2 -1 Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Section 6.3 Lengths of Plane Curves 38 1 17. (a) g=2x^ (I)' =4x^ - L =fY i+ (i0 2dx /^ v 7 ! +4x 2 dx (c) L«6.13 (b) i — -o'.s b 0:5 1 ■ ■ i:j ■ ■ ■ t- ■ 18. (a) g = sec 2 x => (g) 2 = sec 4 x =>■ L = J ^ a/ 1 + sec 4 x dx (c) L«2.06 (b) -] -0.8 -0.6 -0.4 -0.2 19. (a) I = cosy => (^J = cos 2 y => L = I a/1 + cos 2 y dy (c) L « 3.82 (b) 20. (a) dx dy v/i^^y 5 (*)' y l- y 2 L " J-I/2V ! + I 1 ^) X!>-yT 1/2 dy ^JL/S^y (c) L« 1.05 (b) x-vr^F NOT TO SCALE 21. (a) 2y + 2 = 2| => (|) 2 = (y+l) 2 => L = £ iV /l + (y+l) 2 dy (c) L«9.29 (b) y 2 + 2y = 2x + 1 rm-rrr ' Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 382 Chapter 6 Applications of Definite Integrals 22. (a) -Z- = cos x - cos x + x sin x =>- ( ■£■ ] = x 2 sin 2 x => L = J \/l + x2 sin 2 x dx (c) L«4.70 (b) i!> 2.1 (a, I=tanx^ (g)' o (c) L«0.55 L = J \/l +tan 2 x dx = J -S 25 - = sec x dx Jo cos x Jo tt/6 sin^ x + cos^ x (b) dx y OM D2 53 53 5T~ * 24. (a) ! = v /sec 2 y-l => (|) 2 = sec 2 y-l =► L = J^ /3 v/l + (sec 2 y-l) dy Xir/4 r>7r/4 |secy|dy= J secydy (c) L«2.20 (b) x= [ vsec 2 1-1 dt Jo 25. V^x = / o yi+(^) 2 dt,x>0^ v / 2"=yi + (^) 1 real number. dy dx ± 1 => y = f(x) = ± x + C where C is any 26. (a) From the accompanying figure and definition of the differential (change along the tangent line) we see that dy = f'(x k _,) Ax t 4 length of kth tangent fin is V( A x k ) 2 + (dy) 2 = y/( A x k ) 2 + [f'(x k _,) A x k ] 2 . (■h-vf^k.i y=/Cx), Tangent fin with slope A*.,) (b) Length of curve = lim J] (length of kth tangent fin) = lim J] a/( A x k ) 2 + [f'(x k _,) A x k ] 2 k=l k=l = n lim oo £ Vl + [f'(x k -i)] 2 Ax k = / a Vl + [f'W] 2 dx 27. (a) ( jj| ] correspondes to i here, so take ^ as -^-. Then y = y^ + C and since (1, 1) lies on the curve, C = 0. So y = y/x from (1, 1) to (4, 2). (b) Only one. We know the derivative of the function and the value of the function at one value of x. Copyright (c) 1 Pearson E Won, to, publishing as Pearson iii-ltt Section 6.3 Lengths of Plane Curves 383 28. (a) (p-) correspondes to X here, so take ^ as \. Then x = — - + C and, since (0, 1) lies on the curve, C = 1 Soy =T ^. (b) Only one. We know the derivative of the function and the value of the function at one value of x. 29. (a) f = -2sin2tand^f = 2 cos 2t => J (f t ) 2 + (f) = ^{-2 sin 2t) 2 + (2 cos 2t) 2 = 2 => Length = JJ 2 2 dt = [2t] l' 2 = tt (b) f = tt cos 7rt and f -7T Sin 7Tt V S ) 2 + (^) 2 = V 7 ^ COS 7rt)2 + (-7T Sin 7rt)2 = TT =>• Length = J 7r dt = [7rt] _ x , 2 = tt 30. x = a(0 - sin 0) dy ^ °| = a sin =^ ( ^ (8) a(l — cos 0) 2 'dx\ 2 a 2 (1 — 2 cos + cos 2 0) and y = a(l — cos 0) •>2tt a 2 sin 2 => Length = £ * J (|) 2 + (|) d0 = XV 2 ^ 1 - cos 0) d0 V^/r^T 11 ! 2 ^^ = 2a £>in f | d0 = 2a /"sin f d0 = -4a [cos f] J" = 8a 31-36. Example CAS commands: Maple : with( plots ); with( Student[Calculusl] ); with( student ); f := x -> sqrt(l-x A 2);a := -1; b:= 1; N := [2, 4, 8 ]; for n in N do xx := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x,f(x)],x=xx)]; L := simplify(add( distance(pts[i+l],pts[i]), i=l..n )); T := sprintf("#31(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [f(x),pts], x=a..b, title=T ): end do: display( [seq(P[n],n=N)], insequence=true, scaling=constrained ); L := ArcLength( f(x), x=a..b, output=integral ): L = evalf( L ); 37-40. Example CAS commands: Maple : with( plots ); with( student ); x:=t->t A 3/3; y := t -> t A 2/2; a:=0; b:=l; N := [2, 4, 8 ]; for n in N do tt := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x(t),y(t)],t=tt)]; #(b) #(a) #(c) Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle 384 Chapter 6 Applications of Definite Integrals L := simplify(add( student[distance](pts[i+l],pts[i]), i=l..n )); # (b) T := sprintf("#37(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [[x(t),y(t),t=a..b],pts], title=T ): # (a) end do: display( [seq(P[n],n=N)], insequence=true ); ds := t ->sqrt( simplify(D(x)(t) A 2 + D(y)(t) A 2) ): # (c) L := Int( ds(t), t=a..b ): L = evalf(L); 31-40. Example CAS commands: Mathematica : (assigned function and values for a, b, and n may vary) Clear[x, f] {a, b} = {-l, l};f[x_] = Sqrt[l-x 2 ] pl=Plot[f[x], {x, a, b}] n = 8; pts = Table[{xn, f[xn]}, {xn, a, b, (b - a)/n}]// N Show[{pl,Graphics[{Line[pts]}]}] Sum[ Sqrt[ (pts[[i + 1, 1]] - pts[[i, l]]) 2 + (pts[[i + 1, 2]] - pts[[i, 2]]) 2 ], {i, 1, n}] NIntegrate[ Sqrt[ 1 + f [x] 2 ],{x, a, b}] 6.4 MOMENTS AND CENTERS OF MASS 1 . Because the children are balanced, the moment of the system about the origin must be equal to zero: 5 . 80 = x • 100 =^ x = 4 ft, the distance of the 100-lb child from the fulcrum. 2. Suppose the log has length 2a. Align the log along the x-axis so the 100-lb end is placed at x = — a and the x = f . That is, x is located 100(-a) + 200(a) 300 200-lb end at x = a. Then the center of mass x satisfies x at a distance a — I = y = i (2a) which is I of the length of the log from the 200-lb (heavier) end (see figure) 2 or j of the way from the lighter end toward the heavier end. i(2a) 100 lbs. a/3 200 lbs The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point masses located at the centers of the rods at coordinates ( | , 0) and (0, |) . Therefore x = — _ x 1 m 1 +x 2 m2 m l+ m 2 mass location. 5- m +0 m+m j andy yim 2 +y 2 m 2 m!+m 2 0+|-m m+m f j, ¥) is the center of 4. Let the rods have lengths x = L and y = 2L. The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point masses located at the centers of the rods at coordinates (fe, 0) and (0, L). Therefore x |-m+0-2m m+2m !andy=2^n 2L 3 Ji > t) * s tne center of mass location. 5. M = / o 2 x-4dx= [. 4f 4-3 8; M = f\ dx = [4x] 2 =4-2 = 8 Mo _ i M L 6. Mo = J] x • 4 dx = U f 1 = \ (9 - 1) = 16; M = f } 4 dx = [4x]f =12-4 = 8 Mo M 16 8 Copyright (c) 2006 Pearson Education Dt 7. M, o I^ 1 + f)dx /„'( 3 + Mo M :/ x +f)d X =[f+fi; 15_5 9 — 3 Section 6.4 Moments and Centers of Mass 385 f;M = X 3 (i + i)dx=[ x + f '9 , 27 ,2 ' 9 M o=r x (2-t) dx =r( 2x -T)d X =[x 2 16 64 \ 16 -f 16- ?- = ^ Mn M 32 _ 16 3-6 9 ~1T f) 2^3/ 9. M = / i 4 x(l + -l ; )dx = / i 4 (x + X V 2 )d X =[| M = J^l + x- 1 / 2 ) dx = [x + 2X 1 / 2 ] J = (4 + 4) - (1 + 2) = 5 => x=^ = ^ 15 , 14 _ 45+28 _ 73 . 2 ' 3 — 6 — 6 ' 73 30 10. M = J i/4 x - 3 (x- 3 / 2 + X-V2) dx = 3/ i/4 (x-V2 + x -3/ 2) dx = 3 [2x i/2 _ _2_] J /4 = 3 [(2 - 2) - (2 - I U> 3(4 - 1) = 9; M = 3 r 4 (x-3/ 2 + x^/ 2 ) dx = 3 [£ - ^] J = 3 [(-2 - |) - (-4 - f )] = 3 (2+ f) 6 + 14 = 20 1/4 Mo M 20 11. M = J o x(2 - x) dx + Jj x - x dx = f Q (2x - x 2 ) dx + f'x 2 dx = [ = £ = 3; M = J o (2 - x) dx + Jj x dx = 1 2x 2x 2 x 3 2 3 1 1 2- 2 ) + v2 2/ ;i-d + (!-^ M^ M 12. M = / o 'x(x+l)dx + / i 2 2xdx=/ o 1 (x 2 + x)dx + / i 2 2xdx= [f + f ] + [x 2 ]i = (5 + 5) +(4-1) 3 + f = f ; M = / o '(x + 1) dx + \\ dx = [f + x] * + [2x] 2 = (i + l) + (4 - 2) = 2 + § M. ;i)(i; 23 21 13. Since the plate is symmetric about the y-axis and its density is constant, the distribution of mass is symmetric about the y-axis and the center of mass lies on the y-axis. This means that x = 0. It remains to find y = ^ • We model the distribution of mass with vertical strips. The typical strip has center of mass: Cx ,y ) = I x, ^^ ) , length: 4 — x 2 , width: dx, area: dA = (4 — x 2 ) dx, mass: dm = 6 dA = 6 (4 — x 2 ) dx. The moment of the strip about the x-axis is y dm = ( 2L y^ j 6 (4 — x 2 ) dx = | (16 — x 4 ) dx. The moment of the plate about the x-axis is M x = J Ij dm = £j(16-x 4 )dx=f[l6x-^] 2 _ 2 =f[(l6-2 plate is M = / 6 (4 - x 2 ) dx = 6 Ux - f 1 " = 26 (8 0-(-l«-2 + f)] ^P . Therefore y <5-2 Mx M 32 f) !!&•■ The mass of the «! The plate's center of mass is the point (x, y) Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 386 Chapter 6 Applications of Definite Integrals 14. Applying the symmetry argument analogous to the one in Exercise 13, we find x = 0. To find y M we use the y=25-x2 vertical strips technique. The typical strip has center of mass: (x ."y ) = ( x, 25 ~ x j , length: 25 — x 2 , width: dx, area: dA = (25 - x 2 )dx, mass: dm = 8 dA = S (25 - x 2 ) dx. The moment of the strip about the x-axis is 2S 2 ~ x2 J S (25 - x 2 ) dx = | (25 - x 2 ) 2 dx. The moment of the plate about the x-axis is M x = jy dm f_ 5 f (25 - x 2 ) 2 dx = | f s (625 - 50x 2 + x 4 ) dx = g [625x - y dm 2 50 v 3 J- * 3 X -t- 5 50 2 - f (625 • 5 - f ■ 5 3 i) 6 - 625 (5 - f + 1) = 6 ■ 625 - (§) . The mass of the plate is M = J dm = / \. <5 (25 - x 2 ) dx = 6 [25x - 26 15 ^ 8 ■ 5 . Therefore y Mi M ,5.54. (M 3 h\ = 10. The plate's center of mass is the point (x, y) (0, 10). 15. Intersection points: x — x 2 2x x(2 — x) = => x = 0orx = 2. The typical vertical (x-x 2 ) + (-x) N y=x-x 2 y=-x^^^^ 2 strip has center of mass: ( x , y ) = I x, = (x, - y J , length: (x - x 2 ) - (-x) = 2x - x 2 , width: dx, area: dA = (2x — x 2 ) dx, mass: dm = 6 dA = 8 (2x — x 2 ) dx. The moment of the strip about the x-axis is y dm = ( — y J 8 (2x — x 2 ) dx; about the y-axis it isx" dm = x • 8 (2x — x 2 ) dx. Thus, M x = J y dm = - X 2 (I x ' 2 ) ( 2x - x2 ) dx = - iX 2 ( 2x3 - x4 > dx = - f [^ - f ] = - 1 ( 23 - f ) = - 1 - 23 (1 = - f ; M y = / x dm = f\ ■ 8 (2x - x 2 ) dx = ^J^x 2 - x 3 ) = 8 [| x 3 - £| ' = 8 (l 2 * ~ ' -1) S-2 4 12 3 M / dm = £ 8 (2x - x 2 ) dx = sf*(2x - x 2 ) dx = 6 \\ 2 6(4 4h 3 . Therefore, x = ^ \ 3 ) \4Sj 1 and y M, M ¥)(4) (x, y) = (l, — |) is the center of mass. 16. Intersection points: x 2 — 3 = — 2x 2 =>■ 3x 2 — 3 = =^ 3(x - l)(x + 1) = => x = — 1 or x = 1. Applying the symmetry argument analogous to the one in Exercise 13, we find x = 0. The typical vertical strip has center of mass: (x ; y)=(x, - 2 " 2+ f- 3 ) ) = (x,^, length: -2x 2 - (x 2 - 3) = 3 (1 - x 2 ), width: dx, area: dA = 3 (1 — x 2 ) dx, mass: dm = 8 dA = 38 (1 — x 2 ) dx. The moment of the strip about the x-axis is y=x*-3 y dm = \ 6 (-x 2 - 3) (1 - x 2 ) dx = | 8 (x 4 + 3x 2 - x 2 - 3) dx = | 8 (x 4 + 2x 2 - 3) dx; M x = J y dm Sfjx* + 2x 2 - 3) dx = \6 [f + ^ - 3x M = / dm = 3(5^(1 - x 2 ) dx = 38 |x - =>- (x, y ) = (0, — |) is the center of mass. l-*-2(| 38-2(1-1) -3 38 '3+ 10-45 i 3 -V ~~ -'" V 15 46. Therefore, y = 326 . 5 ' 6-32 5-<5-4 Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle Section 6.4 Moments and Centers of Mass 387 17. The typical horizontal strip has center of mass: Cx ^ ) = ( ^-ir-,y ) , length: y — y 3 , width: dy, area: dA = (y — y 3 ) dy, mass: dm = 8 dA = 8 (y — y 3 ) dy. The moment of the strip about the y-axis is x dm < 5 ( I ^)(y-y 3 ) d y = f(y-y 3 ) 2d y (y 2 — 2y 4 + y 6 ) dy; the moment about the x-axis is y dm = <5y (y — y 3 ) dy = 6 (y 2 — y 4 ) dy. Thus, M x = J y dm = 8 I (y 2 — y 4 ) dy = <5 M y = /xdm=fX 1 (y 2 -2y 4 + y«)dy=f[f- 2 f + ^ y 8 f 35 -42 +15 ' 2 V 3-5-7 / hi-i: 28 15 48 105 ; M = / dm *xv dy = 8 \- ii Hl-l) Therefore, x _8_ 15 (x,y) 16 8 is the center of mass. My M" ' 48 ' ,105; 16 105 and y M 1) 18. Intersection points: y = y 2 — y =>■ y 2 — 2y = => y(y — 2) = =>• y = 0ory = 2. The typical horizontal strip has center of mass: (x,y)=(^j) = p,y), length: y — (y 2 — y) = 2y — y 2 , width: dy, area: dA = (2y — y 2 ) dy, mass: dm = 8 dA = 6 (2y — y 2 ) dy. The moment about the y-axis is3c dm = § • y 2 (2y — y 2 ) dy = | (2y 3 — y 4 ) dy; the moment about the x-axis is y 1 dm = <5y (2y — y 2 ) dy = 8 (2y 2 — y 3 ) dy. Thus, M x = Jy dm = I 8(2y 2 - y 3 ) dy = 8 [f - 1 = Tl ( 2 y 3 - y 4 ) d y = I [^ - f ] o = i ( 8 - = 8 [y 2 - f 1 2 = 5 (4 - |) = f . Therefore, x *( ¥) *5 (4 - 3) = f ; M y = / x dm 16-' f( 40_32; 46 . ; M = J dm = f 8 (2y - y 2 ) dy (x,y) M, f4«' v 5 / y4SJ and y , l) is the center of mass. Ma M :f ) (*) 19. Applying the symmetry argument analogous to the one used in Exercise 13, we find x = 0. The typical vertical strip has center of mass: (x ,y ) , length: cos x, width: dx, y=cos x area: dA = cos x dx, mass: dm = 8 dA = 8 cos x dx. The moment of the strip about the x-axis is y dm = 8 ■ ^p • cos x dx | cos 2 x dx 8 ( 1 + cos 2x *\ dx f (1 + cos 2x) dx; thus M , = J y dm = J*'* £ (1 + cos 2x) dx = 8 [sin x if/2 I -tt/2 2 V 2 7T/2 2<5. Therefore, y = ^ -*/2 x + ^1 ' 1 = I [(f + 0) - (- §)] = f ; M = / dm = 8 /* ms x dx i-28 8 2 J -jr/2 20. Applying the symmetry argument analogous to the one used in Exercise 13, we find x = 0. The typical vertical strip has center of mass: ( x ,y ) = I x, ^L* J , length: sec 2 x, width: dx, area: dA = sec 2 x dx, mass: dm = 8 dA = 8 sec 2 x dx. The moment about the x-axis is y* dm = ( ^^ J (8 sec 2 x) dx , y dm = I I . (0, 1) is the center of mass. y /*7r/4 /W 4 sec 4 x dx. M. = | y dm = £ | sec 4 x 5 J-jt/4 ^ 2 J_a-M dx -jt/4 (*> y) km Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 388 Chapter 6 Applications of Definite Integrals /^(tan 2 x + 1) (sec 2 x) dx = § f^\ (tan x) 2 (sec 2 x) dx + f J^sec 2 x dx = f [^ -w/4 I[tanx]t /4 « ri 2 L3 Therefore, y I)] M, }[!-(-!)] ;M= fdm=<5 P sec 2 x dx = <S[tan xf /4 ,, = <5[1 - (-1)] J J —71/4 7T/4 '^ (J_ x \ 3 J V2*/ (x,y) (0, 1) is the center of mass. 26. 21. Since the plate is symmetric about the line x = 1 and its density is constant, the distribution of mass is symmetric about this line and the center of mass lies on it. This means that x = 1. The typical vertical strip has center of mass: t-x 2 )+(2x 2 -4x) (x ,y ) / (2x-x 2 ) + {2x 2 -4x) \ V x ' 2 ) length: (2x - x 2 ) - (2x 2 - 4x) 3x 2 + 6x = 3 (2x - x 2 ) , width: dx, area: dA = 3 (2x — x 2 ) dx, mass: dm = 6 dA = 3<5 (2x — x 2 ) dx. The moment about the x-axis is y dm = | 6 (x 2 - 2x) (2x - x 2 ) dx = - § 6 (x 2 - 2x) 2 dx y = 2x i - Ax - \6 (x 4 - 4x 3 + 4x 2 ) dx. Thus, M, = / y dm = -f o § <5 (x 4 - 4x 3 + 4x 2 ) dx = - § 6 [f - : -^(f-2 4 + 4 .23)=-^.2 4 (|-l + |)=-^.2 4 (^±i0)=-f;M=/dm £36 (2x - x 2 ) dx = 36 \r ■ (x,y) . -: . . v " = 36 (4 - § ) = 46. Therefore, y = f = (- f ) (£) 'l, — I) is the center of mass. 4 3 3 * 22. (a) Since the plate is symmetric about the line x = y and its density is constant, the distribution of mass is symmetric about this line. This means that x = y. The typical vertical strip has center of mass: (x ,y ) = (x, ^p") , length: ^9 - x 2 , width: dx, area: dA = \J 9 — x 2 dx, mass: dm = 6 dA = 6y 9 — x 2 dx. The moment about the x-axis is 2+y2=9 ydm = « (^p^) ^9-x 2 dx = f (9 - x 2 ) dx. Thus, M x = J y dm = £ § (9 - x 2 ) dx = | [9x - y] = § (27 -9) = 96;M = Jdm= J 6 dA = 6 J dA = <5(Area of a quarter of a circle of radius 3) = 6 ( 9 -f) 9ttS 4 Therefore, y = % = (96) (^) = % => (x,y) is the center of mass. (b) Applying the symmetry argument analogous to the one used in Exercise 13, we find that x = 0. The typical vertical strip has the same parameters as in part (a). Thus, M x = fy dm = J" § (9 - x 2 ) dx = 2 £ f (9 - x 2 ) dx = 2(9,5) = 18,5; M = /dm=/5dA = ( 5 /dA = 6(Area of a semi-circle of radius 3) = 6 {^f as in part (a) =^> (x, y) = (0, -) is the center of mass. 2 5sx2+y2=9 - , the same y 7T ' J Copyright (c) 2006 Pearson Education Section 6.4 Moments and Centers of Mass 389 23. Since the plate is symmetric about the line x = y and its density is constant, the distribution of mass is symmetric about this line. This means that x = y. The typical vertical strip has center of mass: (x ,y ) = (x, 3 +V 2 9 ~ x j > length: 3 — v 9 — x 2 , width: dx, area: dA = (3 - a/9 - x 2 ) dx, mass: dm = 6 dA = 6 (3 — v 9 — x 2 j dx. The moment about the x-axis is ^ (3 + N /9^?)(3-v , 9^^) y dm = ^ '- dx = I (4 - 7T) = center of mass. [9 - (9 - x 2 )] dx with side length M = SA = f (4 - tt). Therefore, y <\x- dx. Thus, M, f Jo §[x 3 ] 3 = f . The area equals the area of a square with side length 3 minus one quarter the area of a disk with radius 3 => A = 3 2 — ^ '9S 2 ) I 9(5(4 - w) 2 4-7T (x,y) V4— 7T ' 4 — 7T. is the 24. Applying the symmetry argument analogous to the one used in Exercise 13, we find that y = 0. The typical vertical strip has center of mass: (x ,y ) (x,0), 4 , width: dx, area: dA = A dx, length: £ _ (_ £ mass: dm = 8 dA = ^ dx. The moment about the y-axis is x dm = x • |f dx = | dx. Thus, M y = / x dm = f' J dm = f 2^ dx 26 i]'i 2/> 2<5(a- 1) + 1) a 2 a- 1 - 1 2t5(a-l) a 2a a+1 ; M 1 |dx = 6[-^]; 6(-h (x,y) 2a ^-.0) . Also, lim x = 2. + 1 ' / a — > 00 <5(a 2 -l) Therefore, 25. M x = /ydm=/ i if ■ 5 - ( JO dx = r(^)U 2 )(l)dx=r^dx = 2/;x- 2 dx = 2 [-x" 1 ]? = 2 [(-i) _(_!)]= 2 (I) = 1; M J = /xdm=J i x-«5- (J.) dx = J;x(x 2 )(!)dx = 2j; 2 xdx = 2[f]' = 2(2- ±) =4-1=3; M = /dm = J^<S ( J>) dx = J] x 2 (£) dx = 2 J]' dx = 2[x] 2 = 2(2- 1) = 2. So M — 2 — ' *- X > y^ — V2' 2) V =2/x2 (*.?) • «NH| 3 L 2 x = -rf = I and y = ^ — I =>. (x, y) = (|, ^) is the center of mass 26. We use the vertical strip approach: M x = Jy dm = f^-^ 1 (x - x 2 ) - £ dx = \f {* 2 -x 4 ) - 12xdx = 6£(x 3 -x 5 )dx = 6[^- 6(1-J) M y = Jx dm= J\(x-x 2 )-<Sdx = J o '(x 2 -x 3 )- 12xdx= 12^'(x 3 - x 4 ) dx = 12 [^ - fj = 12 (i - ^ Copffigl (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 390 Chapter 6 Applications of Definite Integrals I2_3 20 5 M /dm=/ (x- X 2 )-^dx=12j;'( X 2 - X 3 )d X =12[f _^]^ = 12(i-i) = {§ = 1. So and y '3 V v5> 2, is the center of mass. ^ 1|dX 224- 3 27. (a) We use the shell method: V = £ 2tt ( r ^J (^ t ) d X = f\nx [4j - (- = 16^/^ d X = 167r/V/ 2 d X = 16tt [| X 3 / 2 ] J = 16jt (§ - 8 - §) = ^= (8 - 1) (b) Since the plate is symmetric about the x-a X is and its density S(x) = - is a function of x alone, the distribution of its mass is symmetric about the x-a X is. This means that y = 0. We use the vertical strip approach to find x: M y = jx dm = J' x • ["-^ - (- 4A 1 • 6 dx = f\ - X - I dx = 8 /V 1 / 2 d X = 8[2 X V^ = 8(2.2-2)=16;M = /dm=j; 4 [^-(^)]^d X = 8/ i 4 (^)(I)d X = 8j;V3/ 2 d X = 8 [-2 X - X / 2 ] * = 8[-l - (-2)] = 8. So x = ^ = f = 2 => (x, y) = (2, 0) is the center of mass, (c) 3 l . 4 \. 4 (2,0) 1 y 4 4 -4 28. (a) We use the disk method: V = / ttR 2 (x) dx = J^ it (4,) dx = 4?r J x~ 2 dx = 4?r [- 1] \ = 4tt [^ - (-1)] = tt[— 1 + 4] = 3tt (b) We model the distribution of mass with vertical strips: M x = J y dm = I ^ - (|) - 6 dx = I ■% - y^c dx = 2/V 3 / 2 dx = 2 |"-=|] = 2[-l - (-2)] = 2; M y = Jx dm = J\ ■ \ ■ 6 dx = 2/V/ 2 dx = 2[^]; = 2[f-f]=f;M=/dm=/ i 42 ^dx = 2/ i 4 f dx = 2/Vv 2d x = 2[2xV^ (c) 2(4 - 2) = 4. So x My M m 4 I and y Ma M 2 _ 1 4 — 2 (x,y) '?, |) is the center of mass. y=2/x ,7i. 29. The mass of a horizontal strip is dm = 6 dA = 6L dy, where L is the width of the triangle at a distance of y above its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have jr = — ^ hy 2 L = I (h - y). Thus, M s = / y dm = f* 6y (g) (h - y) dy = f Jjhy - y 2 ) dy = f [ = / dm =X^(K)(h-y)dy=^r(h-y)dy=^[hy-^] 6b /V _ h^ h \ 2 3 § (h 2 - f ) 5bh 2 (i-i; Sbh 2 6bh 2 Soy M \ 6 y V*bhJ the center of mass lies above the base of the Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Section 6.4 Moments and Centers of Mass 391 triangle one-third of the way toward the opposite vertex. Similarly the other two sides of the triangle can be placed on the x-axis and the same results will occur. Therefore the centroid does lie at the intersection of the medians, as claimed. 30. From the symmetry about the y-axis it follows that x = 0. It also follows that the line through the points (0, 0) and (0, 3) is a median => y = | (3 - 0) = 1 => (x, y) = (0, 1). (0,3) (-1,0) (+1,0) 3 1 . From the symmetry about the line x = y it follows that x = y. It also follows that the line through the points (0, 0) and (|, |) is a median => y = x=|-(|— 0)=| => (x,y)=(I,i). (1,0) 32. From the symmetry about the line x = y it follows that x = y. It also follows that the line through the point (0, 0) and (|,|) is a median 2 (a 3 V2 o) (x,y) v3' 3/ (0,a) (0,0) (a,0) 33. The point of intersection of the median from the vertex (0, b) to the opposite side has coordinates (0, |) => y = (b - 0) - 1 = ' => (x,y)=(f,|). | and x :t-o)-i (0,b) (0,0) (a,0) 34. From the symmetry about the line x = | it follows that x = |. It also follows that the line through the points (f,0) and (f,b) is a median =S> y=I(b-0)=| (x,y) 'a b\ v2' 3) (0,0) (a,0) 35. y = x 1 / 2 =^> dy = \ x" 1 / 2 dx => ds= v/(dx) 2 + (dy) 2 Jo v M, S dx 6 X 2 ^+T dx 2A 3 ;2+i s/2 3/2 f I I X fl\3/2 [(!) 1 + ^dx; l\3/2 , is 3/2 2* 111 _ II 3 V 8 8J 13^ 6 Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 392 Chapter 6 Applications of Definite Integrals 36. y dy = 3x 2 dx =±> dx = J (dx) 2 + (3x 2 dx) 2 = y/l +9x 4 dx; M x = SJ o x 3 Vl +9x 4 dx; [u = 1 + 9x 4 =>■ du = 36x 3 dx =>• ^ du = x 3 dx; x = 0=^u=l,x=l=^u=10] -M x = 5 f , 3LuV 2 du=A [ | u 3/2 ] ;o = ^ (l0 3/ 2 1 y = x- 2kr sin 2 C = 0; 37. From Example 6 we have M s = f a(a sin 0)(k sin 61) d0 = a 2 k£ sin 2 d6» = & J* (1 - cos 20) 69 = &[0- ^M] * = ^ ; M y = JJ a(a cos 0)(k sin 0) d0 = a 2 M — I ak sin d0 = ak[— cos 0]J = 2ak. Therefore, x is the center of mass. M, M 2 k f * sin cos 69 J() -s u Oandy=f = (*=)(£ i n 2ak/ o, 38. M x = J y dm = J" (a sin 9) - <5 - a d0 = J| ) 7r (a 2 sin^)(l + k|cos0|)d0 = a 2 £ (sin 0)(1 + k cos 0) d0 + a 2 r (sin 6»)(1 — k cos 9) 69 *J 7r/2 = a 2 P sin 9 d<9 + a 2 k P sin cos d0 + a 2 f* sin 69 - a 2 k f" sin 9 cos d0 Jo Jo J 7r/2 J 7r/2 a 2 [— cos 0] tt/2 Al^ 1 '" + a 2 [-cos0]J /2 - a 2 k |Mn " 2 [0 - (-1)] + a 2 k (i - 0) + a 2 [-(-l) - 0] - a 2 k (0 - \) k/2 a 2 + ^ + a 2 + ^ = 2a 2 + a 2 k = a 2 (2 + k); M y = Jx dm = J*(a cos 0) • <5 • a d0 = £{a 2 cos 0) (1 + k |cos 0|) d0 = a 2 r (cos 0)(1 + k cos 0) d0 + a 2 f (cos 0)(1 - k cos 0) d0 JO <J 7r/2 = a 2 /J^cos d0 + a 2 k / ( 1+c ° s2fl ) d0 + a 2 f cos d0 - a 2 kjj 2 1 + cos20> d0 a 2 [sin0] o2 t/2 , a 2 k i2»1 7r /2 fa , sin 29 ] "V 2 i .^Tcir, fll^ a 2 k \n , sin2ff l ff « 2 (1 - 0) + & [(I - 0) - (0 + 0)] + a 2 (0 - 1) - & [(tt + 0) - (f + 0)] = a 2 M = J** <5 • a d0 = a J"(l + k |cos 0|) d0 = a /* (1+ k cos 0) d0 + &f* (1 - k cos 0) d0 . 2 k7T „2 a 2 k7T f\. — - a - — - u, tt/2 a[0 + k sin 0]^ + a[0 - k sin 9Y n/2 = a [(f + k) - 0] + a [(tt + 0) - (f - k)] f + ak + a (f + k) = a?r + 2ak = a(7r + 2k). So x (0, 2 T a + 2 k k a ) is the center of mass. and y M, _ a 2 (2 + k) _ a(2 + k) M a(7r + 2k) it + 2k 39. Consider the curve as an infinite number of line segments joined together. From the derivation of arc length we have that the length of a particular segment is ds = i/(dx) 2 + (dy) 2 . This implies that M x = J 6y ds, M y = J 6x ds and M = J 6 ds. If 6 is constant, then x = -^ = ~y s = *-^— and 77 _ m, _ /yds _ /yds ' M J ds length ' 40. Applying the symmetry argument analogous to the one used in Exercise 13, we find that x = 0. The typical Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 393 vertical strip has center of mass: (x ,y ) = I x, —^- J , length: a — |-, width: dx, area: dA = ( a — |- 1 mass: dm = 6 dA = 6 (a - |) dx. Thus, M s = / y dm = f_f\^ \ (a + g) (a. - g) 8 dx dx. SOp- 80-16^1 O 2 -*)*=! [* - *] :, : - 2 • i [ A - *] 7 = s (^ - *W) 2 \/P» 2a 2 6 v /pa 8a-> 2^S ^;M=/dm=* / (a-g) -2^/jS dx 6 ax - 8a<5./pa ^-'■'h-C-"!^^ ) = 4atf ^ (l -±)= 4a6^ (%<*) . Soy M /8aVpaN / 3 N V 5 J \m^i] 3 a, as claimed. 41. Since the density is constant, its value will not affect our answers, so we can set S = 1. A generalization of Example 6 yields M, = J y dm = J a 2 sin dd = a 2 [— cos 3k!° cos (| + a) + cos (| - a)] = a 2 (sin a + sin a) = 2a 2 sin a; M = J dm = J _ "a d0 = alOfJ^l [(I *)] = 2aa. Thus, y Mi M 2a 2 sin a a sin a 2aa Now s = a(2a) and a sin a c = 2a sin a. Then y = 2a" = If > as claimed. 42. (a) First, we note that y = (distance from origin to AB) + d a cos a + d a(sin a — a cos a) Moreover, h = a — a cos a a(sin a — a cos a) sin a — a cos a a(a — a cos a) a — a cos a The graphs below suggest that ZOOM VIEW 0.6 0.5 0.4 0.3 0.2 0.1 ~3$2 3D~ /(<*) = sma - acosa 5 — O.'l o!2 a (b) a 0.2 0.4 0.6 0.8 1.0 f(a) 0.666222 0.664879 0.662615 0.659389 0.655145 6.5 AREAS OF SURFACES OF REVOLUTION AND THE THEOREMS OF PAPPUS 1. (a) |=sec 2 x^ (|) 2 = sec 4 x =>• S = 2n j (tan x) y 1 + sec 4 x dx (c) S « 3.84 (b) ' 0.8 0.6 0.4 y = tanjt 0.2 *' 0.'1 0.'4 6.'6 o:s Copffigl (c| 1 Pearson Educate he, publishing as Pearson Addison-Wesle 394 Chapter 6 Applications of Definite Integrals 2- (a) f x =2^ (%)' = ** => S = 2ir f Q x 2 a/1 +4x 2 dx (c) S « 53.23 (b) dx dy 3. (a) xy = 1 => x = ± =* S = 2^/ i 2 iyiT^dy (c) S « 5.02 fd*\ 2 _ i V d y/ y 1 (b) 03 555 0l7 0.'8 55 f 4. (a) | = cosy => (<|j = cos 2 y =4> S = 27r J (sin y) ^/l + cos 2 y dy (c) S « 14.42 (b) U' 62 53 53 53" r 5. (a) x 1 /2 + y i/2 = 3 ^ y= (3_ x i/2)< > g=2(3-xV 2 )(_I x - 1 / 2) ,2 (S)^ = (1-3 X - 1/2 ) 1 => S = 2tt J]* (3 - x 1 / 2 ) 2 ^/l + (l-3x-!/2) 2 dx (c) S « 63.37 (b) x v J + y i/2 = 3 Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 6. (a) | = l + y-i/2^ (!) 2 = (i + y -i/2) 2 (b) =► S = 2ir f*(y + 200 y/l + (l+ y- 1 ' 2 ) 2 dx (c) S « 51.33 395 7. (a) |=tany => (|) 2 =tan 2 y => S = 2irJ (J tantdt) i/l + tan 2 y dy = 2tt \ I I tan t dt J sec y dy (c) S « 2.08 (b) 3 ' 3.'4 " 3.'8 ' 42 " i'&i'i x IIDJ'B 0:4 0:5 016 0:7 X (a) g = V^I^(g) =x 2 -l (b) => 8 = 2^ 5 (/,Vt 2 - 1 dt) a/1 + (x 2 - 1) dx = 2 -/^(/ 1 X V / ^ r Tdt) xdx (c) S « 8.55 1 1.2 y = I Jt 2 -ldt 1 0.8 0.6 0.4 0.2 \ vi 1:4 \:s i.'j j jj 9- y=f =» g = §;s = />ry</i i- dx => s = />(!) a/TT] TTx/5 4 dx = T^X xdx Try/5 y = 4tt\/ 5; Geometry formula: base circumference = 2-7r(2), slant height = y 4 2 + 2 2 = 2y 5 Lateral surface area = \ (47r) I 2 y 5 ) = 47ry 5 in agreement with the integral value 10. y = § => x = 2y => | = 2; S = J] 2™ ^1 + (|) dy = f o 2tt • 2yy / l+2 2 dy = 4^0 J fl y dy = 2^0 [y 2 ] = 27ry 5 • 4 = 87ry 5; Geometry formula: base circumference = 27r(4), slant height = y 4 2 + 2 2 = 2y 5 =S> Lateral surface area = | (87r) ( 2y 5 ) = 87ry 5 in agreement with the integral value JiV5 Tx 2 ii- | = i;S = />y^i + (g) L dx = j; 3 2 7 rfc±i) v / i + (i) 2 dx=^/;(x+i)dx = ^ [(I + 3) - (I + 1)] = ^ (4 + 2) = 37rv/5; Geometry formula: n = | + | = 1, r 2 = § + | = 2, slant height = y/(2 — l) 2 + (3 — l) 2 = y 5 =>■ Frustum surface area = 7r(rj + r a ) X slant height = 7r(l + 2)y 5 = 37ry 5 in agreement with the integral value Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 396 Chapter 6 Applications of Definite Integrals 12. y I => x = 2y - 1 =4> dx dy 2; S = £l-KX Jl + (|) dy = JjWy - l)x/l + 4 dy = 2ttx/5 £(& - 1) dy = lir^fs [y 2 - y] J = 2tt v/5 [(4 - 2) - (1 - 1)] = 4^^; Geometry formula: n = 1, r 2 = 3, slant height = x/(2 — l) 2 + (3 — l) 2 = \J 5 =>■ Frustum surface area = 7r(l + 3) y 5 = Any 5 in agreement with the integral value 13. dy dx U X = x? _^ (*£\ — X 4 . c _ f~27n 3 =* ^dxj - 9 =^ S - Jo 9 1 + 7 => du = | x 3 dx =>• | du = y dx; 1 + f dx; H / 125 3 V 27 u= l,x = 2 >25/9 1 1 251 u 1 / 2 • i du I \\ u 3 / 2 25/9 7[ ( 1 25-27 ' 3 V 27 / 2 L3 98tt " 81 14. dy _ 1 x -l/2 (*)' 4x 5<lx 27r X" /4 \/^i dx = 27T [f (x 4JT 3 15 4 f (8-1) 1x3/2 4J 28- 3 1x3/2 4) ix 3/21 15/4 ?J J 3/4 f r (|) 3 - 1 y 1.94- ■ 0.87- ■ y=Vx_ 0.75 —i X 3.75 15- t 1 (2 - 2x) 2 \/2x - x 2 1-x \/2x^ (£) : (i-xy 2x — x 2 =► s = X|;W2x-x 2 A /i = 2tt r 1J v / 2x-x 2 V2x ~ x ^± J 0.5 v x/2x = 2 7 r/ o ' 5 5 dx = 27r[x]i;5 = 2tt 2 \/2x — x 2 4- 1 — 2x + x 2 j (1 - x) 2 2x — x 2 X^ dx y=V2x-x 2 — 1 1 1 x .5 1 1.5 dy _ 1 _, ( dy \ ' dx 2v / xTT \ dx / 1 ~ 4(x+l) => S = J^ lllyjx + 1 V 1 + 4(x 1 + l) dx = 27rj;Y(x+l) + i dx = 27Tj r | 5 A /xT f dx =2*[w+\r\\ = f [(5 + I) 3/2 - "(1 + D 3/2 = 4 f[(W /2 -af 2] 4tt /5 3 3 3 \ J — 3 ^23 23 J = |(125-27) = ^ 49-rr 3 y 2.4 ■ ■ 1.4 -■ y=Vx+T 17. dx ._ y 2 dy. - (I) =y 4 ^S = X^^AT7dy; [u = 1 + y 4 =>- du = 4y 3 dy =>■ 1 du = y 3 dy; y = => a = 1, y = 1 => u = 2] -» S = J^tt (i) u 1 / 2 (i du) = l/>du=l[lu 3 / 2 ]; = |(xA-l) Copy |t (c| 1 Pearson Etati, Inc., publishing as Pearson Addison-Wesle Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 397 18. x = (i y 3 / 2 - y 1 / 2 ) < 0, when 1 < y < 3. To get positive area, we take x = — (4 y 3 / 2 — y 1 / 2 ) X= (ly^2_ y1 /2) l (y l/2_ y -l/ 2) ^ (|) = l(y_ 2 + y-l) S = - J> (I y 3 / 2 - y 1 / 2 ) ^1 + j (y - 2 + y-i) dy „3/2_ v l/2\ /I -2tt 4 (y + 2 + y-i)dy / i 3 (i^_ y i/2)^^ dy = _ ff j;V/aQy-l)(yi/» + ^)dy = -7r/ l S (iy-l)(y+l)dy — [(f " ! - 3) - (| - | - 1)] = — (-3 - i + | + 1) -|(-18-l+3) = i|2 19. £ - = 4-7T 15/4 = 4^ => S = / Q 27r-2V4^y0T^dy = 47r/ o ^(4 - y) + 1 dy X' 5/4 v 7 ^ dy = -4tt [f (5 - y) 3 / 2 ] J 5/4 = - & [(5 - ¥) 3/2 - 5 3 / 2 ] = ■ - £ [(f) 3 / 2 _ 5 3 / 2 ^[(5-^) 3/2 -5 3 / 2 ]=-f [ 55; (Sa/s - 5v_ 3 \ J V J g 40^-5^ ^ _ 35tt75 \ _ 8tt A 40^5-575 \ on — — - ZU - dy - y2>^T (I) =2^1 =* S = J^v^T ^1 + ^T dy = 2./ v y (2y - 1) + 1 dy 2 -X>y 1/2 dy = ^ [§ y 3 ^ 2 ] \ /8 = *£ [i 3/2 - (if 2 } = 4 -¥ - $) 21. ds= v/dx 2 + dy 2 y 3 -i) +!dy 4y :; ^(y° - I + w) + i dy = ^(y e + | + i^) dy (y 3 + 40 2 dy = (y 3 + fr) dy; S = J^Try ds = 2^'y (y 3 + ^) dy = 2^(y 4 + \ y- 2 ) dy 2- [? " hi x =2- [(f - |) -(*-*)] =2.(^ + |) = 1(8-31+5)-- 20 22. y = \ (x 2 + 2) 3/2 =>• dy = xyV + 2 dx =>■ ds = a/I + (2x 2 + x 4 ) dx => S = 2tt/ o x v / 1+2x 2 + x 4 dx = 2tt/ o V/5 x v /(x 2 +1) 2 dx = 2 7 r/ o V/5 x (x 2 + 1) dx = 27r/ o "% 3 + x) dx = 2n [£ + f 1 2 = 2tt (| + §) = 4tt 23. y = x/a 2 - x 2 => g = I (a 2 - x 2 )~ 1/2 (-2x) 7* (*)' (a 2 - x 2 ) !> S = 2 7 rJ^ v / a 2 -x 2 jT+ (j^jy dx = 2irf^y/(a 2 - x 2 ) + x 2 dx = 2tt J\ dx = 27ra[x]l a 2?ra[a - (-a)] = (27ra)(2a) = 4?ra 2 24. y dy dx h h ^ \dx) 2jrr h 25. y = cos x — - t - — sin x dy dx = £=► S = 2tt r Jo '^V 1+ P 'h 2 + r 2 [ X 2 ] 2 Jo 27rr ~ h 2 x/h 2 + r 2 (^ (D' = sin 2 X => f.jr/2 S = 2tt , J-jr/2 + H dx Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 398 Chapter 6 Applications of Definite Integrals 26. y = (1 - x*/3) 3/2 => S = | (1 - x 2 / 3 ) 1/2 (- 1 x-V3) = _ (±^ 2 dx 2 .2/3^ 3 / 2 xV3 (*)' 1-x 2 / 3 _ J_ x 2/3 — x 2/3 > S = 2/ o 2tt (1 - x 2 /3) 3/2 ^1 + (-^ - l) dx = 4irf o (l - x 2 / 3 ) 3/2 ^V* dx 47r/ ' (1 - x 2 / 3 ) 3/2 x-V3 dx; [u = 1 - x 2 / 3 => du = - § x" 1 ^ dx - = 0=>u=l,x=l =^ u = 0] -► S = 4tt J" u 3 / 2 (- | du) = -6tt [| u 5 / 2 ] ° = -6tt (0 \ du = x 1/<3 dx; 2\ _ 12tt 5/ — 5 27. The area of the surface of one wok is S = J 2nx J 1 + ( ^ J dy. Now, x 2 + y 2 = 16 2 =4> x = ^/W~^f = *£? ; s = £2-v / T6 2 ^7V 1 + T6^ d y = 2-£>(i6 2 -y 2 ) + y 2 dy dy - ^/162-y 2 "^ = 2tt I 16 dy = 327r • 9 = 2887T « 904.78 cm 2 . The enamel needed to cover one surface of one wok is V = S • 0.5 mm = S • 0.05 cm = (904.78)(0.05) cm 3 = 45.24 cm 3 . For 5000 woks, we need 5000 • V = 5000 • 45.24 cm 3 = (5)(45.24)L = 226.2L => 226.2 liters of each color are needed. 28.y = v / r^ 2 "^ g = - \ -fc (I) =^;s=2./; Vr^^yrr^ Vi' 2 — x 2 vr 2 — x 2 Xa+h _ /»a+h y (r 2 — x 2 ) + x 2 dx = 27rr I dx = 27rrh, which is independent of a. dx 29. y = \/R 2 - x 2 => dy _ _ l 2x . -x dx " 2 y R 2 _ x 2 — ,/ R 2 - x 2 a+h (I) =p^;s = 2./; v^ 2 " x 2 ./l + p^dx = 2nf* v/(R 2 -x 2 )+x 2 dx = 2ttR J & dx = 2?rRh 30. (a) x 2 + y 2 = 45 2 => ^45^7 2 _v dx dy v /452_ y 2 (I) 1 452 _ y 2 S = f 45 2tt a/45 2 J -22.5 v y\n -22.5 v J V ' 45 2 -y 2 = (2?r)(45)(67.5) = 6075tt square feet (b) 19,085 square feet dy = 2n Xl 5 v / (45 2 -y 2 )+y 2 dy = 2tt ■ 45^/y 31. y = x => (|)=1 => (|) =1 => S = 2^/* |x| y/l + 1 dx = 2tt f^(-x)y/2 dx + 2wf\y/2 dx = -2\/27r[f] + 2^/277 [f] = -2\/27r(0- i) + 2 v / 2tt(2-0) = 5^ 32- I = f 33. f |Q = ^ ^ by symmetry of the graph that S = 2 J_°^ 2tt (- f ) ^/l + f dx; [u = 1 + f => du = \ x 3 dx => - 1 du = - f dx; x = -^3 ^u = 2, x = 0^u=lj -> S = 47r/ 2 ' u 1 / 2 (- \) du = — 7T J u 1 / 2 du = — 7T [| u 3//2 ] = — 7T ( | — | \/8 = y ( \/8 — 1 J . If the absolute value bars are dropped the integral for S = J _ 27rf(x) ds will equal zero since I _ 27r I ^- ) J 1 + ^ dx is the integral of an odd function over the symmetric interval — \J 3 < x < y\3. ^ = -sintand| = cost => ^ (^f) 2 + (f )' = V(- sin t) 2 + (cos t) 2 = 1 => S = / 27ry ds = f " 2n(2 + sin t)(l) dt = 2tt [2t - cos t] jf = 27r[(47r - 1) - (0 - 1)] = 8tt 2 Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 34. | = & and | = r^ =» a/(|) 2 + (|) 2 = a/H^ = ^ => S = / 2™ ds 399 J. 2 ^ 2tsm j£+i dt 4£ -V5 / t^t 2 + 1 dt; [u = t 2 + 1 => du = 2t dt; t = => u = 1, [t = a/3 => u = 4J -» J/f A/udu = [f u 3 / 2 ] J :s.. 9 v/3 3/21 J 2+1 dt is an improper integral but lim f(t) exists and is equal to 0, where Note: I 27r(ft f(t) = 2tt (I t 3 / 2 ) v/^ 1 . Thus the discontinuity is removable: define F(t) = f(t) for t > and F(0) = =* Jo F « dt 2S:. 9 ,K 'and & 'V5 35. ^ = land^= . . dtJ (t)' t 2 + 2A/2t+3 =^> S = /2?rxds = /_! 2tt ft + a/2) yt 2 + 2y2t + 3 dt; [u = t 2 + 2-^2 1 + 3 =>• du = ^2t + 2a/2J dt; t = - \/2 => u = 1, t = a/2 => u = 9J -» /Va/u du = [§ 7m 3 / 2 ] j = f (27 - 1) 52^ 3 36 — JU - dt a(l — cos t) and $ = a sin t y(|) 2 +(f) 2 =^[a(l-cost)] 2 + (asint) 2 V a 2 — 2 a 2 cos t + a 2 cos 2 t + a 2 sin 2 t = \J 2a 2 — 2a 2 cos t = ay 2 v 1 — cos t =>- S = J 27ry ds J *2tt a(l - cos t) • a a/2 a/ 1 -cost dt = 2 a/2 tt a 2 J^l - cos t) 3/2 dt 37. | = 2and| = l =► ^f) 2 + (f) 2 = ^^ = ^ => S = J 2ny ds = /J 2^(t + Dy^dt /- r . -I 1 /- /- /- /- = 2-7TV5 I + t = 37rV5. Check: slant height is V 5 => Area is 7r(l + 2)a/5 = 3ir a/5 . 38. £=hand£ dx\ 2 . dt J ) = A/h 2 + r 2 => S = / 2?ry ds = J u 2 7 rrtA/h 2 ~+r 2 d 27rrA/h 2 + r 2 J t dt = 27rrA/h 2 + r 2 l~|| = 7rrA/h 2 + r 2 . Check: slant height is \/h 2 + r Area is rv/h 2 " y=/W 39. (a) An equation of the tangent line segment is (see figure) y = f(m k ) + f'(m k )(x - m k ). When x = x k _i we have ri = f(m k ) + f'(m k )(x t _, - m k ) = f(m k ) + f'(m k ) (- Af) = f(m k ) - f'(m k ) ^ ; when x = x k we have T2 = f(m k ) + f'(m k )(x i - m k ) = f(m k ) + f '(m k ) %* ; (b) L 2 = (Ax k ) 2 + (r 2 - n) 2 = (Ax k ) 2 + [f'(m k ) *? - (-f'(m k ) ^)] 2 = (Ax k ) 2 + [f'(m k )Ax k ] 2 => L k = A/(Ax k ) 2 + [f'(m k )Ax k ] 2 , as claimed (c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent line segment about the x-axis is given by AS k = 7r(ri + r2)Lk = 7r[2f(m k )] y (Ax k )" + [f'(m k )Ax k ] 2 using parts (a) and (b) above. Thus, AS k = 27rf(m k ) a/1 + [f'(m k )] 2 Ax k Copyright (c) 2006 Pearson Education, Inc 400 Chapter 6 Applications of Definite Integrals (d) S = n Km o £ AS t = B Bm ± 2rf(m k ) ^1 + [f (m t )] 2 Ax k = £ 2rf(x) v/l + [f (x)] 2 dx 40. S = J>f(x) dx = I 3 2n- -fc dx = -fc [x 2 ]f = ^ = V^tt ^ UA " n/3 75 41. The centroid of the square is located at (2, 2). The volume is V = (2tt) (y) (A) = (2tt)(2)(8) = 32tt and the surface area is S = (2tt) (y) (L) = (27r)(2) I 4y8 ) = 32\/27r (where \J 8 is the length of a side). 42. The midpoint of the hypotenuse of the triangle is ( | , 3) =4- y = 2x is an equation of the median =4> the line y = 2x contains the centroid. The point ( | , 3) is -¥- units from the origin =>■ the x-coordinate of the centroid solves the equation y (x - -|) 2 + (2x- -3) 2 = 4 => (x 2 -3x+|)+(4x 2 - - 12x + 9) = 5 " 4 =*> 5x 2 - 15x + 9 = -1 =4> x 2 — 3x + 2 = (x — 2)(x — 1) = => x = 1 since the centroid must lie inside the triangle => y = 2. By the Theorem of Pappus, the volume is V = (distance traveled by the centroid)(area of the region) = 2ir (5 — x) [| (3)(6)] = (2tt)(4)(9) = 72tt 43. The centroid is located at (2,0) (2?r) (x) (A) = (27r)(2)(7r) = 4tt 2 44. We create the cone by revolving the triangle with vertices (0, 0), (h, r) and (h, 0) about the x-axis (see the accompanying figure). Thus, the cone has height h and base radius r. By Theorem of Pappus, the lateral surface area swept out by the hypotenuse L is given by S = 2-7ryL = 27T (§) y h 2 + r 2 = 7rrv r 2 + h 2 . To calculate the volume we need the position of the centroid of the triangle. From the diagram we see that (h.r) the centroid lies on the line y = jr x. The x-coordinate of the centroid solves the equation \f (x — h) \^i^^)^~^)^^^ + 4h 2 ) 2h or 4h 3 UI 3 , since the centroid must lie inside the triangle 2h \. By the Theorem of Pappus, V = [2tt fj )] [\ hr) = 1 7rr 2 h. 45. S = 27ry L =>• 47ra 2 = (27ry) (7ra) =>- y = ^, and by symmetry x = 46. S = 27rpL => [2tt (a - f )] (vra) = 27ra 2 (7r - 2) 47. V = 27ryA => | ?rab 2 = (27ry) 48. V = 2^pA => V = [2tt (a + £)] (^ ) 4b y = 5^ and by symmetry x = -a 3 (3-+4) 3 49. V = 2irp A = (27r)(area of the region) • (distance from the centroid to the line y = x — a). We must find the distance from (0, y\ to y = x — a. The line containing the centroid and perpendicular to y = x — a has slope — 1 and contains the point (0, |^) . This line is y = — x + ||. The intersection of y = x — a and y = — x + |Ms the point f4a+3a7r 4a — 3a^^^ (lK 6n Thus, the distance from the centroid to the line y x — a is Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle Section 6.6 Work 401 '4a + 3a7T\ 2 , /4a _ 4a , 3a7r\ 2 _ y/2 (4a + 3a;r) v _ n , / y|(4a + 3a7r) \ /jraA _ 72 ;ra 3 (4 + 3?r) , 6tt J + V3tt 6ir + 6tt J — 6jr ^ V — (Z7T; ^ 6?r y^ 2 y_ fi 50. The line perpendicular to y = x — a and passing through the centroid (0, — ) has equation y = — x + — . The intersection of the two perpendicular lines occurs when x — a = — x + — =>• x = 2a 2 l ^ r a7T y= 2a_^ Thus the distance from the centroid to the line y = x - a is \j ( ^Mp - 0) 2 + ( ^^ - | ) 2 = ^^ Therefore, by the Theorem of Pappus the surface area is S = 2ir (vra) = V27ra 2 (2 + 7r). 5 1 . From Example 4 and Pappus's Theorem for Volumes we have the moment about the x-axis is M x = y M »)(¥) 2ir: 3 6.6 WORK 1 . The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F(x) = kx. The work done by F is W = I F(x) dx = k I x dx = | [x 2 ] =^ k = 400N/m 9k 2 • This work is equal to 1 800 J 1800 ^f = 200 lb/in. 2. (a) We find the force constant from Hooke's Law: F = kx =>• k = - =>• k (b) The work done to stretch the spring 2 inches beyond its natural length is W = I kx dx = 200 J q x dx = 200 |*f 1 = 200(2 - 0) = 400 in • lb = 33.3 ft • lb (c) We substitute F = 1600 into the equation F = 200x to find 1600 = 200x in. 3. We find the force constant from Hooke's law: F = kx. A force of 2 N stretches the spring to 0.02 m ^> 2 = k - (0.02) => k = 100 Si. The force of 4 N will stretch the rubber band y m, where F = ky => y = X0.04 kx dx 4N 100 g r-0.04 100 x Jo dx = 100 5 0.04 (100X0.04) 2 2 0.08 J 4. We find the force constant from Hooke's law: F = kx I => k=2<> => k : 90 — . The work done to stretch the spring 5 m beyond its natural length is W = I kx dx = 90 I x dx = 90 - (90) '25^ 1125 J 5. (a) We find the spring's constant from Hooke's law: F = kx=>k=| = 2LZ11 = 21£li => k = 7238 Pj n0.5 n0.5 (b) The work done to compress the assembly the first half inch is W = I kx dx = 7238 I x dx = 7238 [f 1 = (7238) ^ = < 7238 >< - 25 > « 905 in • lb. The work done to compress the assembly the second half inch is: W = J^ kx dx = 7238 J^ x dx = 7238 [f 1 = ^ [1 - (0.5) (7238)(0.75) 2 2714 in -lb 6. First, we find the force constant from Hooke's law: F = kx => k = - = 41^- = 16 • 150 = 2,400 *. If someone (re) compresses the scale x = I in, he/she must weigh F = kx = 2,400 (|) = 300 lb. The work done to compress the f.l/8 scale this far is W = J o kx dx = 2400 1/8 §^ = 18.75 lb- in. = ff ft - lb Copffigl (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 402 Chapter 6 Applications of Definite Integrals 7. The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to x, the length of the rope still hanging: F(x) = 0.624x. The work done is: W = I F(x) dx = I 0.624x dx = 0.624 \4\ =780 J 8. The weight of sand decreases steadily by 72 lb over the 1 8 ft, at 4 lb/ft. So the weight of sand when the bag is x ft off the ground is F(x) = 144 - 4x. The work done is: W = J F(x) dx = j o (144 - 4x)dx = [144x - 2x 2 ] f = 1944 ft • lb 9. The force required to lift the cable is equal to the weight of the cable paid out: F(x) = (4.5)(180 — x) where x />180 nlSO is the position of the car off the first floor. The work done is: W = / F(x) dx = 4.5 I (180 — x) dx r 2i 180 i [iSOx-V] =4.5( 4.5 180x 4.5 180 2 180 2 4.5-180 2 72,900 ft • lb 10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F(x) t. The work done is W r -^dx kf^dx= k [i]:=k(i k(a - b) ab 11. The force against the piston is F = pA. If V = Ax, where x is the height of the cylinder, then dV = A dx Work = J F dx = J pA dx = J p dV. 12. pV = c, a constant => p Thus W = £ 109,350V- 14 dV = [- ™>] '243 (109,350)(5) (0.4)(36) cV" 1 - 4 . If Vi = 243 in J and pi = 50 lb/in d , then c = (50)(243) 32 243 109,3501b. 109,350 ( 1 v 32 - 4 0.4 243°- 4 ; 109.350 (1 0.4 U -37,968.75 in • lb. Note that when a system is compressed, the work done by the system is negative. 13. Let r = the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to (20 — x), the distance the bucket is being raised. The leakage rate of the water is 0.8 lb/ft raised and the weight of the water in the bucket is F = 0.8(20 — x). So: r 20 r 2i 20 W = J o 0.8(20 - x) dx = 0.8 20x - f I = 160 ft • lb. 14. Let r = the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to (20 — x), the distance the bucket is being raised. The leakage rate of the water is 2 lb/ft raised and the weight of the water in the bucket is F = 2(20 — x). So: r 20 r 21 20 W = J o 2(20 - x) dx = 2 20x - \\ = 400 ft • lb. Note that since the force in Exercise 14 is 2.5 times the force in Exercise 13 at each elevation, the total work is also 2.5 times as great. Copyright (c) 2006 Pearson Education Section 6.6 Work 403 G *ou od 10ft 15. We will use the coordinate system given. (a) The typical slab between the planes at y and y + Ay has a volume of AV = (10)(12) Ay = 120 Ay ft 3 . The force F required to lift the slab is equal to its weight: F = 62.4 AV = 62.4 • 120 Ay lb. The distance through which F must act is about y ft, so the work done lifting the slab is about AW = force x distance = 62.4 - 120 • y • Ay ft • lb. The work it takes to lift all 20 the water is approximately WrsJ] AW 20 = J2 62.4 • 120y • Ay ft • lb. This is a Riemann sum for the function 62.4 • 120y over the interval < y < 20. The work of pumping the tank empty is the limit of these sums: n20 r 21 20 W = J o 62.4 • 120y dy = (62.4)(120) I \ I = (62.4)(120) (*f) = (62.4)(120)(200) -hp motor is t w 250 fHb 1,497,600 ft-lb 250 Mb 1,497,600 ft -lb 5990.4 sec (b) The time t it takes to empty the full tank with ( -^ = 1 .664 hr =4> t « 1 hr and 40 min (c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is W = J o 62.4 • 120y dy = (62.4)(120) NU = (62.4)(120) (±f) = 374,400 ft • lb and the time is t = 1497.6 sec = 0.416 hr w 25 min (d) In a location where water weighs 62.26 — • 250 ^ ft 3 ■ a) W = (62.26)(24,000) = 1,494,240 ft • lb. b)t 1,494,240 250 5976.96 sec w 1.660 hr In a location where water weighs 62.59 IV" a) W = (62.59)(24,000) = 1,502,160 ft - lb b)t 1,502,160 250 6008.64 sec w 1.669 hr 1 hr and 40 min 1 hr and 40.1 min 16. We will use the coordinate system given. "Trr^^ .Ground level (a) The typical slab between the planes at y and y + Ay has a volume of AV = (20)(12) Ay = 240 Ay ft 3 . The force F required to lift the slab is equal to its weight: F = 62.4 AV = 62.4 • 240 Ay lb. The distance through which F must act is about y ft, so the work done lifting the slab is about AW = force x distance 20 = 62.4 • 240 • y • Ay ft • lb. The work it takes to lift all the water is approximately W w J2 AW 10 20 = ^2 62.4 • 240y • Ay ft • lb. This is a Riemann sum for the function 62.4 • 240y over the interval 10 />20 10 < y < 20. The work it takes to empty the cistern is the limit of these sums: W = 1 62.4 • 240y dy 2l) (b) t (62.4)(240) I \ I = (62.4)(240)(200 - 50) = (62.4)(240)(150) = 2,246,400 ft • lb 2,246,400 ft-lb 275 W 275 ! 8168.73 sec w 2.27 hours w 2 hr and 16.1 min (c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is W = J^62.4 • 240y dy = (62.4)(240) [y| n = (62.4)(240) (^ - ±f) = (62.4)(240) ' 125 ^ 936,000 ft. Then the time is t \Y 275 ! 936,000 275 . J 10 3403.64 sec 56.7 min Copyright (c) 2006 Pearson Education 404 Chapter 6 Applications of Definite Integrals (d) In a location where water weighs 62.26 — • A» ■ a) W = (62.26)(240)(150) = 2,241,360 ft - lb. b)t 2,241,360 275 8150.40 sec = 2.264 hours « 2 hr and 15.8 min 933,900 c) W = (62.26)(240) (±f ) = 933,900 ft • lb; t - , 7 , In a location where water weighs 62.59 p a) W = (62.59)(240)(150) = 2,253,240 ft - lb. b) t = 2 ' 25 2 jf° = 8193.60 sec = 2.276 hours ra 2 hr and 16.56 min c) W = (62.59)(240) (^) = 938,850 ft • lb; t = ^^ « 3414 sec 3396 sec « 0.94 hours rj 56.6 min 0.95 hours ss 56.9 min 17. The slab is a disk of area 7rx thickness Ay, and height below the top of the tank (10 — y). So the work to pump the oil in this slab, AW, is 57(10 — y)7r(|) . The work to pump all the oil to the top of the tank is W = X'^CLOy 2 - y 3 )dy = ^ U^f- - £1 = 11,875tt ft • lb » 37,306 ft • lb. 18. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is (14 — y)(7r) (|) and since the tank is half full and the volume of the original cone is V = ^7rr 2 h = i7r(5 2 )(10) with half the volume the cone is filled to a height y. ■j 'in. G 1 v 2 3^4y^y ^ ft 3 , half the volume 7500, 250?r fi-3 ' 500 ft. SoW /; ft\ and [ (14y 2 -y 3 ) dy 577T |~ 14y 3 4 3 : r.( ii ) 60,042 ft • lb. 19. The typical slab between the planes at y and and y + Ay has a volume of AV = 7r(radius) 2 (thickness) '20\ 2 Ay = 7T • 100 Ay ft 3 . The force F required to lift the slab is equal to its weight: F = 51.2 AV = 51.2 • 1007T Ay lb =^> F = 51207T Ay lb. The distance through which F must act is about (30 — y) ft. The work it takes to lift all the 30 30 kerosene is approximately W w Y AW = Y 51207r(30 — y) Ay ft • lb which is a Riemann sum. The work to pump the tank dry is the limit of these sums: W= I 51207r(30 — y) dy = 51207T 30y w 7,238,229.48 ft • lb ::ii 5120^(250) (5120)(450tt) 20. (Alternate Solution) Each method must pump all of the water the 15 ft to the base of the tank. Pumping to the rim requires all the water to be pumped an additional 6 feet. Pumping into the bottom requires that the water be pumped an average of 3 additional feet. Thus pumping through the valve requires y 3 ft(47r)6 ft 3 (62.4 lb/ft 3 ) « 14,1 15 ft • lb less work and thus less time. 21. (a) Follow all the steps of Example 5 but make the substitution of 64.5 p for 57 p. Then, W X o S^s: (10 - y)y 2 dy 64.5tt ioy 4 3 64.5 4 E ( 'lOJ V 3 ' 64.5 v 4 -) (8 3 ) -2 64.5;r-i 3 21.5tt-8 3 « 34,582.65 ft • lb (b) Exactly as done in Example 5 but change the distance through which F acts to distance w (13 — y) ft. Then W = p Q Z* (13 - y)y 2 dy = ^ [ = (19tt) (8 2 ) (7)(2) w 53.482.5 ft - lb 13y 3 3 r] = 5J r{ 1J f~^) = ( 5J r)(S 3 )( l i-2) 57tt-& j -7 3-4 22. The typical slab between the planes of y and y+Ay has a volume of about AV = 7r(radius) 2 (thickness) = 7T (^/y) 2 Ay = xy Ay m 3 . The force F(y) is equal to the slab's weight: F(y) = 10,000 ^ • AV = 7iT0,000y Ay N. The height of the tank is 4 2 = 16 m. The distance through which F(y) must act to lift the slab to the level of the top of the tank is about (16 — y) m, so the work done lifting the slab is about Copyright (c| 1 Pearson Etation, Inc., publishing as Pearson Addison-Wesle Section 6.6 Work 405 AW = 10,0007ry(16 — y) Ay N • m. The work done lifting all the slabs from y = 0toy= 16 to the top is 16 approximately W w Yl 10,0007ry(16 — y)Ay. Taking the limit of these Riemann sums, we get o 16 W = J 10,0007ry(16 - y) dy = 10,000tt/ (16y - y 2 ) dy = 10,000tt 21,446,605.9 J 16y 2 2 10,000tt ( 16 3 16 3 1 0.000- tt-16 3 6 23. The typical slab between the planes at y and y+Ay has a volume of about AV = 7r(radius) 2 (thickness) = 7r (a/25 — y 2 ) Ay m 3 . The force F(y) required to lift this slab is equal to its weight: F(y) = 9800 • AV = 9800tt (a/25 -y 2 ) 2 Ay = 9800?r (25 - y 2 ) Ay N. The distance through which F(y) must act to lift the slab to the level of 4 m above the top of the reservoir is about (4 — y) m, so the work done is approximately AW « 98007T (25 - y 2 ) (4 - y) Ay N • m. The work done lifting all the slabs from y = -5 m to y = m is approximately W rs ]T] 98007T (25 — y 2 ) (4 — y) Ay N • m. Taking the limit of these Riemann sums, we get -5 W = J_° 5 98007r (25 - y 2 ) (4 - y) dy = 9800tt /_° 5 (100 - 25y - 4y 2 + y 3 ) dy = 9800tt [lOOj/ - f y 2 - \ y 3 = -9800tt (-500 - ^ + § • 125 + ^f) * 15,073,099.75 J 24. The typical slab between the planes at y and y+Ay has a volume of about AV = 7r(radius) (thickness) = n (a/100 -y 2 ) 2 Ay = tt (100 - y 2 ) Ay ft 3 . The force is F(y) = ^ • AV = 56?r (100 - y 2 ) Ay lb. The distance through which F(y) must act to lift the slab to the level of 2 ft above the top of the tank is about (12 - y) ft, so the work done is AW « 567T (100 - y 2 ) (12 - y) Ay lb - ft. The work done lifting all the slabs 10 from y = ft to y = 10 ft is approximately W « Yl 567r ( 100 — y 2 ) (12 — y) Ay lb • ft. Taking the limit of these Riemann sums, we get W = f 56tt (100 - y 2 ) (12 - y) dy = 56ttJ o (100 - y 2 ) (12 - y) dy = 567r/ o 10 (1200 - lOOy - 12y 2 + y 3 ) dy = 56tt [l200y 1 00y- 2 nf 3 10 56tt (12,000 10,000 4 - 1000 + ISfiQO) (56tt)(12-5-4- . .„„.,, . , ,,._ .. , .,, (1000) w 967,611 ft -lb. It would cost (0.5)(967,61 1) = 483,8050 = $4838.05. Yes, you can afford to hire the firm. 25. F = m dv mv p- by the chain rule W /: mv fr dx = m /: dv i dx = m [i v 2 (x \ m [v 2 (x2) — v 2 (xi)] = | mv| — \ mv 2 , as claimed. 26. weight = 2 oz = ^ lb; mass = "^ = 55 = 555 slu § s ; W = (1/ 27. 90mph W 90 mi 1 hr 1 min 5280 ft 1 hr 60 min 60 sec 1 mi l)(iiii)( 132ft/sec ) 2 « 85 - lft - lb 132 ft/sec; m = |^ (255 slugs) (160 ft/sec) 2 w 50 ft- lb °-^ slugs; 28. weight = 1.6oz = 0.1 lb 0.1 lb 32 ft/sec 2 L slugs; W 320 3I0 slugs) (280 ft/sec) 2 = 122.5 ft • lb 29. weight = 2 oz W I lb m = A slugs 1 slugs; 124 mph = (1 f fi 4 ,^ 0) w 181.87 ft/sec; 25 ft (60)(60) i) (^ slugs) (181.87 ft/sec) 2 « 64.6 ft • lb 30. weight = 14.5 oz 14=5 lb 16 1U 14.5 (16)(32) slugs; W = (i) ((i^) slugs j (88 ft/sec) 2 w 109.7 ft ■ lb 31. weight = 6.5 oz = j| lb =$> m 6.5 (16)(32) slugs; W = (1) ((1H2) slu S s ) ( 132 ft/sec) 2 « 110.6 ft • j Copffigl (c) 1 Pearson M^ he, publishing as Pearson Addison-Wesle 406 Chapter 6 Applications of Definite Integrals 32. F = (18 lb/ft)x => W = J* 18x dx = [9x 2 ] J /6 = \ ft - lb. Now W = \ mv 2 - 1 mv 2 , where W = 5 ft- lb, 32 ~~ 256 slugs and vi = ft/sec. Thus, 4 ft • lb ,2J V256 slugs) v 2 =>■ v = 8 V 2 ft/sec. With v = at the top of the bearing's path and v = 8 y 2 — 32t =>• t = ■ 2 4- sec when the bearing is at the top of its path. The height the bearing reaches is s = 8 y 2 1 — 16t 2 8v / 2)(^)-(16)(^) 2 = 2ft att the bearing reaches a height of 33. (a) From the diagram, r(y) = 60 - x = 60 - y^O 2 - (y - 325) 2 for 325 < y < 375 ft. (b) The volume of a horizontal slice of the funnel is AV«7r[r(y)] 2 Ay 60- v '50 2 -(y-325)^ Ay (c) The work required to lift the single slice of water is AW sa 62.4AV(375 - y) 62.4(375 - y)7r 60 50 2 - (y - 325) 5 Ay. The total work to pump our the funnel is W fSJ5 J 325 62.4(375 - y)7r 6.3358- 10 7 ft ■ lb. 60 50 2 - (y - 325) dy y-325 34. (a) From the result in Example 6, the work to pump out the throat is 1,353,869,354 ft • lb. Therefor, the total work required to pump out the throat and the funnel is 1,353,869,354 + 63,358,000 = 1,417227,354 ft ■ lb. (b) In horsepower-hours, the work required to pump out the glory hole is ' { ^ 1( ^r =715.8. Therefore, it would take 7 i„ 8 h h p p h = 0.7158 hours « 43 minutes. 35. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0, 7]. The typical slab between the planes at y and y + Ay has a volume of about AV = 7r(radius) 2 (thickness) = 7r >y+VL5\ 2 14 Ay in 3 . The force F(y) required to lift this slab is equal to its weight: F(y) = | AV = y ( y+ 14 ' ) Ay oz. The distance through which F(y) must act to lift this slab to the level of 1 inch above the top is about (8 — y) in. The work done lifting the slab is about AW= ¥ 4tt\ (y+17.5) : 42^- (8 — y) Ay in • oz. The work done lifting all the slabs from y = to y = 7 is approximately W = Yl 9^52 (y + 17.5) 2 (8 — y) Ay in • oz which is a Riemann sum. The work is the limit of these sums as the norm of the partition goes to zero: W = I y^jp (y + 17.5) 2 (8 — y) dy ¥w f ( 2450 - 26 - 25 y - 2? y 2 - y 3 ) 4y=fo\-4- 9y 3 - 2J f y 2 + 2451 )>< 4tt _ Zi _ Q 73 _ 26.25 7 2 9-142 4 y ' ' 2 ' ' 9-14- [ 4 2450 71 w 91.32 in • oz 36. We fill the pipe and the tank. To find the work required to fill the tank follow Example 6 with radius = 10 ft. Then AV = 7T • 100 Ay ft 3 . The force required will be F = 62.4 • AV = 62.4 • 100tt Ay = 6240?r Ay lb. The distance through which F must act is y so the work done lifting the slab is about AWi = 62407T • y - Ay lb - ft. The work it takes to Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle Section 6.7 Fluid Pressures and Forces 407 385 385 lift all the water into the tank is: Wi « J2 AWi = J2 62407T - y • Ay lb • ft. Taking the limit we end up with 360 360 - 385 r ,1 385 „,„ = ^ [385 2 - 360 2 ] 360 2 To find the work required to fill the pipe, do as above, but take the radius to be | in = g ft. Then AV = n • A Ay ft 3 and F = 62.4 • AV Wi = J 36o 62407ry dy = 6240tt |"f j 182,557,949 ft -lb ^j 21 Ay. Also take different limits of summation and integration: W2 ~ J2 AW2 Wo r Jo 62.4 36 7ry dy 62.4tt yf 36 2 :!6(i 1 62Air ' v 36 . (¥) 352,864 ft • lb. The total work is W = Wi + W 2 « 182,557,949 + 352,864 w 182,910,813 ft • lb. The time it takes to fill the tank and the pipe is Time = j^ 182.910.813 1650 110,855 sec w 31 hr 37. Work = i^o^MG dr = 1000M G J 6,370,000 r^ J 6 ; 37C = (1000) (5.975 - 10 24 ) (6.672 • lO" 11 ) (^ dr i/w\a*/~i r n 35.780.000 1000 MG [- ?] 6,370,000 35,780,000 5.144 x 10 10 J 38. (a) Let p be the x-coordinate of the second electron. Then r 2 = (p — l) 2 W J°_F(p) dp 1: (23xlQ- 29 ) i i/'-ii J dp ~ L /■-' j-i 23xl0" 29 (23 x 10" 29 ) (1 11.5 x 10 -29 (b) W = Wi + W2 where Wi is the work done against the field of the first electron and W2 is the work done against the field of the second electron. Let p be the x-coordinate of the third electron. Then r 2 = (p — l) 2 and r 2 = (p + I) 2 =* Wl = X 5 ^ dp = J 3 5 f^ dp = -23 x 10- 29 [ 29 I 1 (-23 x 10" 29 ) 23 x 10 -29 and W9 -23xl0- 29 U^ 4 " - > '2 = (-23 x 10- 29 ) (1 - i) 1: 23xl0" 29 ^1^^ ^=^ (3 - 2) = Q x 10- 29 . Therefore 12 W = W x + W 2 = (f x 10- 29 ) + (| x 10- 29 ) f x 10- 29 7.67 x 10 -29 6.7 FLUID PRESSURES AND FORCES 1 . To find the width of the plate at a typical depth y, we first find an equation for the line of the plate's right-hand edge: y = x — 5. If we let x denote the width of the right-hand half of the triangle at depth y, then x = 5 + y and the total width is L(y) = 2x = 2(5 + y). The depth of the strip is (— y). The force exerted by the water against one side of the plate is therefore F = I w(— y) • L(y) dy = I 62.4 • (— y) - 2(5 + y) dy = 124.8 J_" 5 2 (-5y - y 2 ) dy = 124.8 [- § y 2 - \ y 3 ] ~j = 124.8 [(- f - 4 + \ ■ 8) - (- f • 25 + \ ■ 125)] = (124.8) (V - W- y 2 ) dy = 124.8 [- | y 2 ) = (124.8) {^m-- 1684.8 lb 2. An equation for the line of the plate's right-hand edge is y = x — 3 =S> x=y + 3. Thus the total width is L(y) = 2x = 2(y + 3). The depth of the strip is (2 — y). The force exerted by the water is F = J\w(2 - y)L(y) dy = J^62.4 • (2 - y) • 2(3 + y) dy = 124.8/^(6 - y - y 2 ) dy = 124.8 [6y - £ - ^ = (-124.8) (-18- |+9) = (-124.8) (-f) = 1684.81b 3. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge is y = x — 3 => x = y + 3. Thus the total width is L(y) = 2x = 2(y + 3). The depth of the strip changes to (4 — y) =>" F = /I w < 4 - y)L(y) dy = J_° 3 62.4 - (4 - y) - 2(y + 3) dy = 124.8 J_° 3 (12 + y - y 2 ) dy = 124.8 [l2y + t-£] = (-124.8) (-36 + § + 9) = (-124.8) (- f ) = 2808 lb Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 408 Chapter 6 Applications of Definite Integrals Using the coordinate system of Exercise 4, we see that the equation for the line of the plate's right-hand edge remains the same: y = x — 3 => x = 3 + y and L(y) = 2x = 2(y + 3). The depth of the strip changes to (— y) =* F = X.° 3 w(-y)L(y) dy = J°62.4 • (-y) • 2(y + 3) dy = 124.8/j-y 2 - 3y) dy = 124.8 [- y - } = (-124.8) (f - |) = (-'W7X2-3) = 561 61b If -3 5. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge to be 2x y + 4 and L(y) = 2x = y + 4. The depth of the strip is (1 — y). (a) F = J_° 4 w(l - y)L(y) dy = J^62.4 • (1 - y)(y + 4) dy = 62.4 fjA - 3y - y 2 ) dy = 62.4 Uy - ^ - f = (-62.4) [(-4)(4) - (b) F = (-64.0) [(-4)(4) - (3X16) _ 2 (3)(16) f) + 64 (-62.4) (-16 -24 _ (-64.0X-120 + 64) _ H94 7 lb (-62.4X-120 + 64) 3 1164.81b 6. Using the coordinate system given, we find an equation for the line of the plate's right-hand edge to be y = — 2x + 4 => x = ^=2 and L(y) = 2x = 4 - y. The depth of the strip is (1 — y) =>- F= I w(l— y)(4 — y) dy 62.4^' (y 2 - 5y + 4) dy = 62.4 [y - ^ + 4y (62.4) 4) = (62.4) '2-15 + 24> (62.4X11) 114.41b y(ft) 7. Using the coordinate system given in the accompanying figure, we see that the total width is L(y) = 63 and the depth of the strip is (33.5 - y) =>• F = / w(33.5 - y)L(y) dy J o 33 ^ - (33.5 - y) - 63 dy = (§) (63)/" (33.5 - y) dy (§) (63) f33.5y - ' 64-63 ^ v 123 J (33.5)(33) - ^ (64)(63)(33)(67 - 33) (2) (123) 13091b y(in) 33.51 surface "33!ol x(in) -31.5 31. S (a) Use the coordinate system given in the accompanying figure. The depth of the strip is ( ^ — y) ft => F =/" /6 w (if -y) (width) dy (62.4)(width)/ (62.4)(width) f ' ' (62.4)(width) [(f / 6 . 6 y 2 j » F end = (62.4)(2) ( l -§) (1) (b) Use the coordinate system given in the accompanying figure. Find Y from the condition that the entire volume of the water is conserved (no spilling): 4i - 2 - 4 = 2 - 2 - Y => Y = y ft. The depth of a typical strip is ( T - y) ft and the total width is L(y) = 2 ft. Thus, F = I" 3w (T-y) L oo d y 11/3 y(ft) i I sur face WWW* W ''y^yy&ffiy&yy ■■:■■:■■:■ -l Wmm'mmm x(ft) 209.73 lb and F side = (62.4)(4) ($) (1) I surface x(ft) / "(62.4) (f - y) • 2dy = (62.4)(2) [f y - f ] " = (62.4)(2) [(1 :^) 2 419.47 lb (62.4)(121) 838.93 lb the fluid force doubles. Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Section 6.7 Fluid Pressures and Forces 409 Using the coordinate system given in the accompanying figure, we see that the right-hand edge is x = y/l — y 2 so the total width is L(y) = 2x = 2yl — y 2 and the depth of the strip is (— y). The force exerted by the water is therefore F = I w • (— y) • 2yl — y 2 dy 62.4/^1 - y 2 d(l-y 2 )= 62.4 [§(1 ,2\3A x(ft) (62.4) (I) (1 -0) = 416 lb 10. Using the same coordinate system as in Exercise 15, the right-hand edge is x = ^/3 2 — y 2 and the total width is / 2 . The depth of the strip is (— y). The force exerted by the milk is therefore L(y) = 2x = 2^/9 F = J° 3 w - (-y) • 2 a/9 - y 2 dy = 64.5 /_°y 9 - y 2 d (9 - y 2 ) = 64.5 [§ (9 - y = (64.5)(18) = 11611b 2x3/2 -3 (64.5) (?) (27-0) 11. The coordinate system is given in the text. The right-hand edge is x = */y and the total width is L(y) = 2x = 2,/y. (a) The depth of the strip is (2 — y) so the force exerted by the liquid on the gate is F = I w(2 — y)L(y) dy = J o '50(2 - y) • 20 dy = 100 £(2 - y)0 dy = lOo/J (2y J / 2 - y 3/ 2 ) d y = 100 [f y 3 / 2 - | y 5 / 2 ] J 100 I (20 - 6) = 93.33 lb (b) We need to solve 160 = / w(H — y) • 2^/y dy for h. 160 100 (f H = 3 ft. 12. Use the coordinate system given in the accompanying figure. The total width is L(y) = 1. (a) The depth of the strip is (3 — 1) — y = (2 — y) ft. The force exerted by the fluid in the window is F = £ W (2 - y)L(y) dy = 62.4 £(2 - y) • 1 dy = (62.4) fey - ^1 - <^> ^ (i - ¥\ - <^M (b) Suppose that H is the maximum height to which the tank can be filled without exceeding its design limitation. This means that the depth of a typical strip is (H — 1) — y and the force is F = j a W [(H - 1) - y]L(y) dy = F nlM , where F max = 312 lb. Thus, F max = w£ [(H - 1) - y] • 1 dy = (62.4) |"(H - l)y 93.6 lb x(ft) (^) (2H - 3) = -93.6 + 62.4H. Then F max = -93.6 + 62.4H 6.5 ft 312 = -93.6 + 62.4H H 405.6 62.4 13. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for j y. The total width is L(y) = 2x = | y and the the line of the end plate's right-hand edge is y depth of the typical horizontal strip at level y is (h — y). Then the force is F = I w(h — y)L(y) dy = F max , where F max = 6667 lb. Hence, F raax = w£ (h - y) • | y dy = (62.4) (f ) / (hy - y 2 ) dy " hy 2 (62.4) (I) Height = h and \ (Base) = I h i] o = (62.4) (I) (f - f ) = (62.4) (I) (I) h 3 = (10.4) (*) h 3 => h = \/(f) (ffej) f water which the tank can hold is V = | (B h 2 ) (30) = 12h 2 « 12(9.288) 2 « 1035 ft 3 . !) (Wz) ~ 9 ' 288 ft The volume of water which the tank can hold is V = \ (Base)(Height) • 30, where V '1u2\ Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 410 Chapter 6 Applications of Definite Integrals 14. (a) After 9 hours of filling there are V = 1000 • 9 = 9000 cubic feet of water in the pool. The level of the water h = jJHj = 6 ft. The depth of the typical horizontal strip at is h = ^-, where Area = 50 • 30 = 1500 Area' level y is then (6 — y) for the coordinate system given in the text. An equation for the drain plate's right-hand edge is y = x =>• total width is L(y) = 2x = 2y. Thus the force against the drain plate is 2-1 (62.4)(2) | \ F=J g w(6 - y)L(y) dy = 62.4 J q (6 - y) • 2y dy = (62.4)(2)/ q (6y - y = (124.8) (3 - i) = (124.8) (f ) = 332.8 lb (b) Suppose that h is the maximum height. Then, the depth of a typical strip is (h — y) and the force F = J w(h - y)L(y) dy = F, llilx , where F max = 520 lb. Hence, F max = (62.4) J* (h - y) • 2y dy 124.8 J o '(hy - y 2 ) dy = (124.8) [ hy 2 y J ~2 T (124.8) (| - i) = (20.8)(3h - 2) 520 20.8 3h-2 h = f = 9 f t 15. The pressure at level y is p(y) = w • y => the average pressure is p = ± j Q p(y) dy = \ J o w • y dy = 1 w j^J = (?) (V) = ^t ■ ^ n ' s ' s ^ e P ressure at level | , which is the pressure at the middle of the plate. 16. The force exerted by the fluid is F = I w(depth)(length) dy = I w • y • a dy = (w • a) I y dy = (w • a) y = w ( 2 ) = (if) ( a ^) = P ' Area, where p is the average value of the pressure (see Exercise 21). /o (62.4) (2y 4 — y 2 ) (— y) dy (62.4) f° (4 - " 2N i 1/2 ''- ■>-'■ .i> - "'->"> i . I = , i , j , ' - y 2 ) 1/z (-2y) dy = (62.4) \i (4 - y 2 ) (62.4) (|) (4 3 / 2 ) = 332.8 ft - lb. The force 3V- 'J j_ 2 -v— ->\m compressing the spring is F = lOOx, so when the tank is full we have 332.8 = lOOx => x w 3.33 ft. Therefore the movable end does not reach the required 5 ft to allow drainage => the tank will overflow. 18. (a) Using the given coordinate system we see that the total width is L(y) = 3 and the depth of the strip is (3 — y). Thus, F = £ w(3 - y)L(y) dy = J fl 3 (62.4)(3 - y) • 3 dy = (62.4)(3)/ o 3 (3 - y) dy = (62.4)(3) [3y - £] ' = (62.4)(3) (9 - |) = (62.4)(3) (§) = 842.4 lb •x(tt) -1.5 1.5 (b) Find a new water level Y such that F Y = (0.75)(842.4 lb) = 631.8 lb. The new depth of the strip is (Y — y) and Y is the new upper limit of integration. Thus, F Y = / w(Y — y)L(y) dy 62A£(Y - y) • 3 dy = (62.4X3)]^ (Y - y) dy = (62.4)(3) [Yy (62.4)(3) (?f\ . Therefore, Y = (62.4)(3) ( Y 2 - S 2F Y (62.4X3) 1263.6 187.2 v/6T75 w 2.598 ft. So, AY = 3 - Y 3 - 2.598 rs 0.402 ft « 4.8 in 19. Use a coordinate system with y = at the bottom of the carton and with L(y) = 3.75 and the depth of a typical strip being (7.75 - y). Then F = /" 5 w(7.75 - V)Hy) dy = ( 6 -§) (3.75)_ j f" 5 (7.75 - y) dy 1 64.5 > V 123 ) (3.75) |7.75y- \ 123 (3.75) 4.21b Copyright (c) 2006 Pearson Education Chapter 6 Practice Exercises 411 20. The force against the base is F base = pA = whA = w • h • (length)(width) = (^) (10)(5.75)(3.5) « 6.64 lb. To find the fluid force against each side, use a coordinate system with y = at the bottom of the can, so that the depth of a w(10 — y) ' width of \ . the side / dy .§)(££) iQy-i 57 \ /width of N / 100 \ 12 3 J V the side M 2 / 2L) (50)(3.5) « 5.773 lb and F side = (S) (50)(5.75) « 9.484 lb 123; 21. (a) An equation of the right-hand edge is y y and L(y) = 2x il The depth of the strip is (3 - y) =► F = J\(3 - y)L(y) dy = J 3 (62.4)(3 - y) (f y) dy = (62.4) • (£ J> - y 2 ) dy (62.4) (fn |y- 3 3 „2 (62.4) (I) [f - f ] = (62.4) (I) (f ) = 374.4 lb (b) We want to find a new water level Y such that F Y = \ (374.4) = 187.2 lb. The new depth of the strip is (Y — y), and Y is the new upper limit of integration. Thus, Fy = I w(Y — y)L(y) dy 62.4^ Y (Y - y) (| y) dy = (62.4) (fjJjYy - y 2 ) dy = (62.4) (f) [y 5l _ y_ 2 3 (62.4) (1) (? - f ) (62.4) (§) Y 3 . Therefore Y 3 9Fi (9)(187.2) 3 / (9)(187.2) Vl3.5 « 2.3811 ft. So, v9 y ^ . i^x.^xw x 2.(62.4) 124.8 ~^ \J 124.8 AY = 3-Y«3- 2.381 1 « 0.6189 ft « 7.5 in. to the nearest half inch, (c) No, it does not matter how long the trough is. The fluid pressure and the resulting force depend only on depth of the water. 22. The area of a strip of the face of height Ay and parallel to the base is 100(||) • Ay, where the factor of || accounts for the inclination of the face of the dam. With the origin at the bottom of the dam, the force on the face is then: F = /, 24 w(24 - y)(100)(ff)dy = 6760 [24y - £ 1 ** = 676o(24 2 - %f\ = 1,946,8801b. CHAPTER 6 PRACTICE EXERCISES 1. A(x) = 1 (diameter) 2 = 5( X -2 v /^-x 2 - Hv/x-x 2 0,b= 1 V = J A(x) dx = I J o (x - 2x 5 /' 2 + x 4 dx |x 7 / 2 + 4 ^2 % (35 - 40 + 14) 4-70 280 y=Sx 2. A(x) = 1 (side) 2 (sin |) 73~ ■lyfi. TZ (4x - 4xa/x + x 2 ) ; a = 0, b = 4 V = J a A(x) dx = &f o (4x - 4x 3 / 2 + x 2 ) dx [2x 2 - f x 5 / 2 4 73 4 32^3 4 75 ^(32-^ 32 | 64 \ 5^(15-24+10)= ^ Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 412 Chapter 6 Applications of Definite Integrals 3. A(x) = | (diameter) 2 = | (2 sin x - 2 cos x) 2 = | • 4 (sin 2 x — 2 sin x cos x + cos 2 x) = tt(1 -sin2x);a= f , b = ^ A(x)dx = 7r (1 -sin2x)dx J-tt/4 -[(f+^)-(f-^)]=- 2 4. A(x) = (edge) 2 = N v/6- v/xV-Oj = (v^ - ,/xV = 36 - 24^/6 y'xH- 36x - 4^/6 x 3 / 2 + x : a = 0,b = 6 ^ V = £a(\) dx = £ ^36 - 24^ y/x. + 36x - 4^/6 x 3 / 2 + x 2 ) dx = [36x - 24v/6 • § x 3 / 2 + 18x 2 - \\fl ■ \ x 5 / 2 + f] = 216 - 16 • v^ v^ - 6 + 18 • 6 2 - § v^ v^ - 6 = 216 - 576 + 648 - ^ + 72 = 360 - ^ = inoo-ms = n 2 _,_ §! 5. A(x) = ^ (diameter) 2 = f (2-v/x - f ) = f (4x - x 5 / 2 + fg ) ; a = 0, b = 4 =4> V = /^(x) dx [ J o 4 ( 4 x-x5/ 2 + ^)dx=f [2x 2 -fx^ 2 + ^] o =f(32- 32-f + | - 32) ^(i-f+i; |f (35 - 40 + 14) = ^ 6. A(x) = \ (edge) 2 sin (f ) = f [2^x - {-2y/x)] /l ^ 4,/x) 2 = 4\/3x; a = 0, b = 1 >. V = J^AOO dx = f*4y/3 x dx = [2-^3 x 2 ] 2^ y 2 = 4x 7. (a) rfisfc method: V = _£ b 7rR 2 (x) dx = J\ (3x 4 ) 2 dx = tt /V 8 dx = TT [X 9 ] _j = 2TT (b) sfoeZZ method: V = J> (** ) (** ) dx = J>x (3*») dx = 2, - 3 JV dx = 27T - 3 [f ] * = tt Note: The lower limit of integration is rather than — 1 . (c) shell method: v=r 2 -(^)(^) dx = 2 ^ i " x )( 3x4 ) dx = 2 - [¥-Ci =27r til -i) -(-§-1)]=^ (d) washer method: R(x) = 3, r(x) = 3 - 3x 4 = 3 (1 - x 4 ) => V = f n [R 2 (x) - r 2 (x)] dx = J_ |7 r [9 — 9(1 — x 4 ) = 9tt Jjl - (1 - 2x 4 + x 8 )] dx = 9tt J_' 1 (2x 4 - x 8 ) dx = 9n \ dx 2x° ?r 5 9 [2 _ 11 _ 2tt-13 _ 267T L5 9J 5 5 Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle ;d 2 V=/ a ^[R 2 (x)-r 2 (x)]dx = / i ^[(4 s ) : '-3)]=^(-i7)-5 + T + ?)=l»(-2-10 + 64 + 5) dx = 2tt -4x _1 M(- (a) washer method: R(x) = 4j , r(x) = \ = *[($-*)-■ (b) s/ieZZ method: (c) s/ieZZ method: V = ar^CSl) (** ) dx = 2 -r^ 2 " x ) (P - 5) dx = 2»jT(i = 2tt [- I + f -x + f ] 2 = 2tt [(-1 + 2 - 2+ 1) - (-4 + 4 - 1 (d) washer method: ? V = / tt[R 2 (x) - r 2 (x)] dx Chapter 6 Practice Exercises dx = 7T [- f X- 5 413 x] 2 4 J 1 57- 20 I)} 2- (I) 5jr 2 -£-1 dx 11 - 16ttJ] (1 -2x -3 , v -6\ dx 49?r 4 49?r 4 49jr 4 49jr 4 167T X + X" -2 X" 16tt[(2- 16tt [far 160 '1 (40 1 _ " 4 1 160 5-32^ 4) -(1 + 1-1)] 1 + 32) 49.-; 4 7_hr 10 103_7T 20 2- x (a) rfisfc method: V = 7T J] (a/x- l) dx = 7T J] (X - 1) dX = 7T [f - xl = *[(¥-5)-(£-l)]=*(¥-4)=8* (b) washer method: R(y) = 5, r(y) = y 2 + 1 => V = £ tt [R 2 (y) - r 2 (y)] dy = tt £ [25 - (y 2 + l) 2 ] dy = njj25 - y 4 - 2y 2 - 1) dy = tt £(24 - y 4 - 2y 2 ) dy = tt [24y - £ - f y 3 ] ^ = 2tt (24 - 2 - f - f = 32tt (3 - § - i) = ^ (45 - 6 - 5) = ^ (c) rfisfc method: R(Y) = 5 - (y 2 + 1) = 4 - y 2 => V = J ttR 2 (y) dy = f_n (4 - y 2 ) 2 dy = tt J^(16-8y 2 + y 4 )dy = 7r[l6y-f + fl 2 =27r(32-f + f ) 64tt (1 - § + 5) = ^ (15 -10 + 3) 5127T 15 10. (a) shell method: v = £ 2, (** ) (*J) dy = />y (y - J) dy = 2 -r(y 2 ^) d y= 2 -[T-^;= 2 -(f-f) 2- _ . 64 = ^ 12 UH 3 Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 414 Chapter 6 Applications of Definite Integrals (b) shell method: V = J> ( ** ) ( **) dx = />x ( 2v /x - x) dx = 2. £(2x 3/2 ~ x 2 ) dx = 2n [| x^ - f ] * = 27r(|-32-f) = if£ (c) s/ieZZ method: v = j> (it ) (it) dx = r^ 4 - *) ( 2 v^ - x ) dx = 2 -x> xi/2 - 4x - 2 * 3/2 + x2 ) dx = 2ir [f x 3 / 2 - 2x 2 - § x 5 / 2 + f ] 4 = 2^ (4 , ,-32-f-32+f)=647r(f-l-f + §) 64ir (1 - |) = %?■ (d) sfoeZZ method: V = r2.(^»j( h t;: 1 i ; t )dy = />(4-y)(y-^)dy = 2.X 4 (4y-y 2 -y 2 + ^)dy = 27r/ o 4 (4y - 2y 2 + {) dy = 2^ [2y 2 - f y 3 + g] * = 2tt (32 - § - 64 + 16) = 32tt (2- f + l) 32- 3 11. disk method: tan J x dx = 7r I (sec x — 1) dx = 7r[tan x — x] tt/3 _ t(3V3-7tj 12. disfc method: V = 7rJ (2 - sin x) 2 dx = 7T J (4-4 sin x + sin 2 x) dx = ttJ (4-4 sin x + '~ c ° s2x ) dx = tt [4x + 4 cos x + | - 22j2s] * = 7T [(4tt - 4 + f - 0) - (0 + 4 + - 0)] = tt (f - 8) = f (9tt - 16) 13. (a) disk method: 32 V = tt/ o (x 2 - 2x) 2 dx = 7T J o (x 4 - 4x 3 + 4x 2 ) dx = tt [^ - x 4 + f x 3 j = tt v , 16+ f) ^ (6 - 15 + 10) = ^ 2 _ 8tt 5 — 5 (b) washer method: V = / Vl 2 - (x 2 - 2x + l) 2 j dx = /Vdx + f\ (x - l) 4 dx = 2tt - [tt &=£] = 2tt - (c) shell method: V = /> (■» (fiSt) dx = 2^(2 - x) [- (x 2 - 2x)] dx = 2^(2 - x) (2x - x 2 ) dx = 2tt£(4x - 2x 2 - 2x 2 + x 3 ) dx = 2tt J^(x 3 - 4x 2 + 4x) dx = 2n H - f x 3 + 2x 2 j = 2tt (4 - f + 8 = f (36 - 32) = f (d) washer method: V = tt Jj2 - (x 2 - 2x)] 2 dx - tt£ 2 2 dx = tt £ U-4 (x 2 - 2x) + (x 2 - 2x) 2 j dx - 8tt = nf (4 - 4x 2 + 8x + x 4 - 4x 3 + 4x 2 ) dx - 8tt = tt J q (x 4 - 4x 3 + 8x + 4) dx - 8tt tt | ^ - x 4 + 4x 2 + 4xT - 8tt = tt (f - 16 + 16 + 8) - 8tt = f (32 + 40) - 8tt = ^ - ^ = ^ 14. disfc method: "ir/4 V = 2?r JJ 4 tan 2 x dx = 8tt J* (sec 2 x - 1) dx = 8?r[tan x - x]q /4 = 2tt(4 - tt) Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Chapter 6 Practice Exercises 415 15. The material removed from the sphere consists of a cylinder and two "caps." From the diagram, the height of the cylinder is 2h, where h 2 2 2 ,i.e. h= l.Thus V cy i = (2h)7r( y3 j = 6tt ft 3 . To get the volume of a cap, use the disk method and x 2 + y 2 = 2 2 : V cap = I 7rx 2 dy = / i 2 7r(4-y 2 )dy = ^[4y-^] i 2 = tt[(8 - f) - (4- i)] = r f ft 3 . Therefore, * removed — * cyl T" ^ V ca p — D7T 12ZL — 28tt f t 3 3 — 3 ii . 16. We rotate the region enclosed by the curve y = J 12 (1 4x2n 121/ and the x-axis around the x-axis. To find the volume we use the disk method: V = J ^ 7rR 2 (x) dx = J t: (v/12 (l - y§|) J dx = 7r J \2 M - yfM dx = 12 -lt( 1 -^) dX=12 "[ X 4x J 363 247T |^ ' 4 \ / JJ_\ 3 V363/ 1 Ui '32., !-(&)(£) 1327T 1 264t 3 3tt w 276 in 3 17- y = xV 2 -^^ | = i x -i/2_I xl / 2 ^ (|) =|(^-2 + x) =.L=/ i 7l + H^-2 + x)dx => L = J i 4 v /l(I + 2 + x)dx = J^ (X-1/2 + x l/2) 2 dx = J" I ( x -l/2 + x l/2) dx = I [ 2x l/2 + 2 x 3/2] J = |[(4+l-8)-(2+|)]=i(2+f) 3 18. x = y 2 / 3 => ^ = lx- 1 /3 dy (!)'■ 4x~ 2 / 3 9 L =XV 1+ (I) * = I V 1 + ^ * r \/9xV3 + 4 3xV3 dx = 5/V9X 2 / 3 + 4 (x- 1 / 3 ) dx; [u = 9x 2 / 3 +4 => du = ey" 1 / 3 dy; x = 1 => u = 13, X = 8 => U = 40] -+ L = i £° U V2 du = _L [| u 3/2] JO = J. [ 40 3/2 _ ^3/2] M 7^34 19. y = 5. x 6/5 _ 5 x 4/5 => g = 1 x 1 / 5 - 1 X-VB =* (g)" = I (x 2 / 5 - 2 + X" 2 / 5 ) => L=/ Jl + i(x 2 / 5 -2 + x- 2 / 5 )dx => L = / Ji (x 2 / 5 + 2 + x- 2 / 5 ) dx = / v/± ( X V5 + x -!/5) 2 dx j; 32 I(x 1 / 5 +x-V 5 )dx=I[^ 6/5 + ix^]f=I[(f.2« + |.2^)-(f + 5)] = I(^ + -) 1(1260 + 450)=^ = ^ 48 20. x=iy 3 + i => | = |y 2 -^ (I) =i^y 4 ^ + ^L=/Vl + ( 1 Ly^I + ^)dy f;^ 6 y 4 + ^^y = fJ{ly 2 + ^~«y = L{h 2 + ^«y={hy 3 -l] 1 (3.- i) _ (1.-1) = 2. + i - 13 V 12 2^ v 12 ^ 12 T 5 12 21. f t = -5 sin t + 5 sin 5t and f 5 cos t — 5 cos 5t , dt) (*)' (—5 sin t + 5 sin 5t) + (5 cos t — 5 cos 5t) 5 v sin 2 5t — 2sin t sin 5t + sin 2 1 + cos 2 1 — 2cos t cos 5t + cos 2 5t = 5 \Jl — 2(sin t sin 5t + cos t cos 5 t) 5^/2(1 - cos it) = 5 J 4(\) (1 -cos4t) = lO^sin 2 2t = 10|sin 2t| = lOsin 2t (since < t < | ) Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 416 Chapter 6 Applications of Definite Integrals Length = J^ 10sin2tdt = [-5cos2t]o /2 = (-5)(-l) - (-5)(1) = 10 22. | =3t 2 - 12tand^ 3t 2 + 12t^ J(f) + (if J = y / (3t 2 -12t) 2 + (3t 2 + 12t) 2 = ^28812 + 18t 4 3 v / 2|tl\A6+l 2 ^ Length = J Q 3^ |t|-/l6 + t 2 dt = 3^^ t v / 16+l 2 dt; [u = 16 + t 2 =>• du = 2t dt =* idu = t dt; t = => u = 16; t = 1 => u = l?j ; ^ / ^du = ^ [|u 3/2 ] JJ = ^ (§(17) 3/2 - f (16) 3/2 h£ . §((17) 3/2 - 64) = ^l({ll) m - 64) 8.617. 23. % = -3 sin and & :! cos w = 4 / ( £ ) - + ($j\ = y (-3 sin (9) 2 + (3 cos 6») 2 = v/3(sin 2 61 3ir/2 r-3ir/2 cos 2 (?) />3ir/2 /.37T/2 Length =J o 3d0 = 3j o &6 = 3(f - 0) 2 24. x = t 2 and y = f - t, -a/3 < t < \/3 =4> 2t and dy •n/3 ■n/3 t 2 - 1 => Length •V3 J-V3 •V5 (2t) / + (t 2 -l) z dt 4x/3 ^3 -V3 25. Intersection points: 3 — x 2 = 2x 2 =>• 3x 2 — 3 = =>- 3(x - l)(x + 1) = =4> x = -1 orx = 1. Symmetry suggests that x = 0. The typical vertical strip has center of mass: (x $ ) = ( x, + ( 2 ~ x - J = ( x, 2L -±^ J , length: (3 - x 2 ) - 2x 2 = 3 (1 - x 2 ), width: dx, area: dA = 3(1— x 2 ) dx, and mass: dm = 6 ■ dA = 36 (1 — x 2 ) dx =4> the moment about the x-axis is y dm= |<S(x 2 + 3)(l-x 2 )dx= \8 (-x 4 - 2x 2 + 3) dx => M x = / y dm = \ 6 J_ t (-x 4 - 2x 2 + 3) dx l4-f-¥+3x 35 (- i - I + 3) = fS (-3 - 10 + 45) 35 x 5 3 66(1 - i) =45 ^ y 3^ 15 32.^ 5 ,M= Jdm= 35/^(1 -x 2 )dx Ml = |M _ £ _ Therefore, the centroid is (x,y) = (0, f ) 26. Symmetry suggests that x = 0. The typical vertical strip has center of mass: ( x , y ) = (x, f ) , length: x 2 , width: dx, area: dA = x 2 dx, mass: dm = S - dA = <5x 2 dx s 2 the moment about the x-axis is y dm = | x 2 • x 2 dx x 4 dx =>• M x = / y dm = § f x : dx=^[x^ 26 (o5\ 10 V 4 i 32A M /dm = 5 J_; 2 x 2 dx = 6[f] f(2 3 ) y = m = tm = s ■ Therefore, the centroid is(x, y) = (0, Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Chapter 6 Practice Exercises 417 27. The typical vertical strip has: center of mass: (x , y ) 4+^\ , ... .2 y=(l/4)x2 x, -^f- j , length: 4 — \, width: dx, area: dA = (4 — ^- )dx, mass: dm = 5 • dA = 6 (4 — ^ ] dx => the moment about the x-axis is y dm = 6 • ^^ (4 - f ) dx = § (l6 - fg) dx; the moment about the y-axis is 3c dm = <5 ( 4 — ^ I • x dx = <5 ( 4x — ^ J dx. Thus, M x = J y 1 dm = § I ( 16 — h ) dx = f[ 1 6x-5 5 1 6]^ = f[64-f]=^;M y = /xdm = ^X 4 (4x-f)dx=^[2x2-fl]^ = 5(32- 16)= 166; M = J dm = 6 £U- f) dx = 6 Ux- f^l = 6 (16 64\ 12 J 325 3 M " 16-6-3 " 32-5 ~ = j and y = M x M 128-5-3 " 5-32-5 12 Therefore, the centroid is (x, y) '3 m v2' 5 ) 28. A typical horizontal strip has: center of mass: (3c , y 1 ) = ( y %~ y , y ) , length: 2y — y 2 , width: dy, area: dA = (2y — y 2 ) dy, mass: dm = 6 ■ dA = S (2y — y 2 ) dy; the moment about the x-axis is y dm = 6 - y • (2y — y 2 ) dy = S (2y 2 — y 3 ) ; the moment about the y-axis is 3c dm = 6 - ^ y | y ^ • (2y — y 2 ) dy = f (4y 2 - y 4 ) dy => M x = / y dm = 6 Jj2y 2 - y 3 ) dy 2 „3 y 4 l _ i. (2 o 16 5 (±8 _ 32-1 2 V 3 5 ) nt-s-f *(¥-¥) 5-16 12 -•M 325 M J"3cdm=fJ>y 2 -y 4 )dy = f [ Jdm = 4 2 (2y-y 2 )dy = 5 [y 2 - ^ = £ (4 - § 45 3 M. M 4-5-3 3-4-5 1. Therefore, the centroid is (x, y) = (|, l) . My M 4 y 3_ 5-32-3 15-5-4 and 29. A typical horizontal strip has: center of mass: (3c , y ) = f^ir^y) . len g th: 2 y - y 2 > width: dy, area: dA = (2y — y 2 ) dy, mass: dm = 8 ■ dA = (1 + y) (2y — y 2 ) dy =>- the moment about the x-axis isy dm = y(l + y) (2y — y 2 ) dy = (2y 2 + 2y 3 - y 3 - y 4 ) dy = (2y 2 + y 3 — y 4 ) dy; the moment about the y-axis is 3c dm = (^) (1 + y) (2y - y 2 ) dy = \ (Ay 2 - y 4 ) (1 + y) dy = \ (4y 2 + 4y 3 y 4 - y 5 ) dy M / y dm = J^(2y 2 + y 3 - y 4 ) dy = [f 2 y 3 ]6 i ]6 J "t" A w, i6 a + |§ (20 +15 -24) =£(11) 44. M 15 ' lvl y l( 4 #+2 4 T(2y + y 2 - (S)(i)H 5 f)=4(| 4 1 9 _ 4 _ 8\ z 5 b) /3c dm = £\ (4y 2 + 4y 3 - y 4 - y 5 ) dy = \ [f y 3 + y 4 - £ - f ; M = /dm = / o 2 (1 + y) (2y - y 2 ) dy y 3 )dy (4- 4(2- J) T> *=% = &) fandy=^ |4 . Therefore, the center of mass is (x, y) '9 n \ v5' 10J 30. A typical vertical strip has: center of mass: (3c , y ) area: dA 3 dx, mass: dm = 8 - dA ^■2 dx *' 2xV2. the moment about the x-axis is , length: -Ik , width: dx, X J /2 ' Copfiigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 418 Chapter 6 Applications of Definite Integrals 'y dm = ^4/2 ' b-m dx = Pj dx; the moment about the y-axis is x" dm = x • tf-fU dx = SrL dx. 2x 3/2 W x 3/2 •>9 x'- 1 (a) M x = 5 j;i(|)dx=f [-^] i = ^;M y = 4 i x(^)dx = 35[2xV^ = 126; M = ^ 9 ^dx=^[x-V 2 ]; =4 ^x=t = ^ = 3andy=^=(l)=f (b) M x = jj (|) dx = § [- I] J = 4; M y = J> (£) dx = [2^] J = 52; M = £ x , = 6[xV^ = 12^ J .= ^ = fandy=t = | x" - dx 3i. s = £2«ysJi + (%) dx;| = ^TT =* (I) =2^1 ^S = X 3 2 7 r^/2xTT^l + 2l ^ T dx = 27r/ 3 v^ + T yM dx = 2^/ o 3 v^TT dx = 2^ [§ (x + l) 3 / 2 ] I = ly/li: . § (8 - 1) = ^2 3ZS = / i 2«y^l+(g) d*;|=x»=>(g) = x* =► S = J* o 2tt - f v^T^dx = |/ o v'TT^ (4x3) = I X l ^/^T^d(l +x^) = 1 [f (1 + x 4 ) 3/2 ] ^ = f [2V2- 1] dx . S = />x^/l + (|) dy dx _ (|)(4-2y) _ 2-y dy ' ^4y^2 " ,/4y~^ >2 (I) 1 4y - y 2 + 4 - 4y + y 2 4y_y2 4y-r dy = 47r I dx = 47r 4y-y 2 34. S = />x^l + (I) dy; | = ^ =* 1 + (|) = = ^/ 2 6 y47TTdy = f [f (4y + l) 3 / 2 ] J = f (125 - 27) = | (98) J_ _ 4y+l 4y — 4y T 2 Vy- v / 4y+T x/4y dy -IS).r 3 35. x = f and y = 2t, < t < y/l => f t = t and ^ = 2 => Surface Area = J o 5 27r(2t) v / t 2 + 4 dt = J]' 27TU 1 / 2 du 2. [§ U 3/ 2 ] ; 76»r where u = t 2 + 4 => du = 2t dt; t = => u = 4, t = x/5 => u 36. x = t 2 + I and y = 4^, 4- < t < 1 =* f = 2t - ^ and $ Surface Area = J ' 2ir (t 2 + i 2t-^) 2 + ($) * = 2 */. M v ;2t+^dt = 2 - O t2 + r t ) (» + & dt = 2, j;^ +1+1 1 -3) dt = 2. [i t* + 1 1 - 1 1 - 2 ] ; /V5 = 2.(2-^) 37. The equipment alone: the force required to lift the equipment is equal to its weight => Fi(x) = 100 N. Xb />40 Fi(x) dx = I 100 dx = [100x]q° = 4000 J; the rope alone: the force required to lift the rope is equal to the weight of the rope paid out at elevation x =>• F2(x) = 0.8(40 — x). The work Xb /*40 r F 2 (x)dx=J () 0.8(40 -x)dx = 0.8 40x the total work is W = W 1 + W 2 = 4000 + 640 = 4640 J 40 0.8 (40 2 - f ) (0.8X1600) 640 J; 38. The force required to lift the water is equal to the water's weight, which varies steadily from 8 • 800 lb to 8 • 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight is F(x) = 8 - 800 - ^ 2-4750 - x x 2-4750 (6400) (1 9500 lb. The work done is W = I F(x) dx Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle L Jo 22,800,000 ft • lb 6400 (1- g&j) dx 6400 x- 2-9500 J 47o() 6400 4750 4750 2 4-4750 Chapter 6 Practice Exercises ) = (|) (6400)(4750) 419 39. Force constant: F = kx =>■ 20 = W = J o kx dx = k f o x dx = |20 k- 1 1 > k = 20 lb/ft; the work to stretch the spring 1 ft is 10 ft - lb; the work to stretch the spring an additional foot is W = f* kx dx = k f\ dx = 20 [f ] \ = 20 ( I |) 20 i 30 ft • lb 40. Force constant: F = kx =>■ 200 = k(0.8) =4> k = 250 N/m; the 300 N force stretches the spring x = | = |^g = 1.2 m; the work required to stretch the spring that far is then W = I F(x) dx = I 250x dx = [125x 2 ]i- 2 = 125(1.2) 2 = 180 J 41. We imagine the water divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0, 8]. The typical slab between the planes at y and y + Ay has a volume of about AV = 7r(radius) 2 (thickness) =& y 2 Ay ft 3 . The force F(y) required to = t(!y) Ay lift this slab is equal to its weight: F(y) = 62.4 AV _ (62.4X25) 7ry2 Ay lb The distance through which F(y) must act to lift this slab to the level 6 ft above the top is about (6 + 8 — y) ft, so the work done lifting the slab is about AW y 14 8 /^ x 4 " -10 10 Reservoir's Cross Section (62.4)(25) 16 7ry 2 (14 - y) Ay ft • lb. The work done lifting all the slabs from y = to y = 8 to the level 6 ft above the top is approximately W«E (62.4)(25) [(S 7ry (14 — y) Ay ft • lb so the work to pump the water is the limit of these Riemann sums as the norm of the partition goes to zero: W = I ( ' )( - /o (16) Vd4-y)dy (62.4) (^f) (t • g3 ~ f ) ~ 418,208.81 ft • lb (62^25), ^(^2 y 3 )dy = (62.4)(^)[fy 3 -£] 42. The same as in Exercise 41, but change the distance through which F(y) must act to (8 — y) rather than (6 + 8 — y). Also change the upper limit of integration from 8 to 5. The integral is: w = | o 5 «^p y2(8 _ y) dy = (62 4) (2^) ^ (8y2 _ y3) dy = (624) (25.) J 8 y 3 _ t] * = (62.4)(^)(f.5 3 -?) 54,241.56 ft -lb 43. The tank's cross section looks like the figure in Exercise 41 with right edge given by x = 4, y = ?. A typical horizontal slab has volume AV = 7r(radius) 2 (thickness) 'Ay I y 2 Ay. The force required to lift this slab is its weight: F(y) = 60 • | y 2 Ay. The distance through which F(y) must act is (2 + 10 — y) ft, so the work to pump the liquid is W = 60 J o tt(12 - y) (^) dy = 15tt [^ - £1 = 22,500tt ft • lb; the time needed to empty the tank is f^°°™ 257 sec 44. A typical horizontal slab has volume about AV = (20)(2x)Ay = (20) (2a/16 — y 2 ) Ay and the force required to lift this slab is its weight F(y) = (57)(20) (2-^/16 — y 2 ) Ay. The distance through which F(y) must act is (6 + 4 — y) ft, so the work to pump the olive oil from the half-full tank is W = 57/_° 4 (10 - y)(20) (2^16 - y 2 ) dy = 2880 f_ 4 10^16 - y 2 dy + 114o/_° 4 (16 - y 2 ) 1/2 (-2y) dy Copffigl (c) 1 Pearson Etoatio^, k, publishing as Pearson Addison-Wesle 420 Chapter 6 Applications of Definite Integrals = 22,800 • (area of a quarter circle having radius 4) + \ (1 140) [(16 - y 2 ) 3/2 j = (22,800)(4tt) + 48,640 335,153.25 ft -lb 45. F = £w • ( d s ^ t p h ) • L(y) dy => F = 2 £(62.4X2 - y)(2y) dy = 249.6£(2y - y 2 ) dy = 249.6 [y 2 _ r (249.6) (4 - f) = (249.6) (f) = 332.8 lb 46. F = JV - (**) - L(y) dy => F = £ 6 ?5 (| - y) (2y + 4) dy = 75 £ '(§ y + f - 2y 2 - 4y) dy 75 X 5/ °(f-ly- 2 y 2 )dy = 75[fy-^ 2 - 2 y3]f = (75)[(f§)-(|)(|)-(f)(l)] (75)(3075) (75) 25 _ 175 _ 250 9 216 3-216^ V9-2 ^g) (25 -216 -175-9-250-3)- „..,,„ k6^ \36J \3J \216J 118.631b. 47. F = JV - ( d ^ t p h ) • L(y) dy =* F = 62.4^ (9 - y) (2 - ^) dy = 62a£ (9y 1 / 2 - 3y 3 / 2 ) dy = 62.4 [6y 3 / 2 - I f' 2 } J = (62.4) (6 • 8 - § • 32) = (^) (48 - 5 - 64) = ^^2 = 2196.48 lb 48. Place the origin at the bottom of the tank. Then F = I W • ( d ^ n t p h J • L(y) dy, h = the height of the mercury column, strip depth = h - y, L(y) = 1 =► F = £849(11 - y) 1 dy = (849) f g (h - y) dy = 849 [hy - f 1 = 849 (h 2 - f ) = ^h 2 . Now solve ^h 2 = 40000 to get h w 9.707 ft. The volume of the mercury is s 2 h = l 2 • 9.707 = 9.707 ft 3 . F = wi/ o 6 (8 - y)(2)(6 - y) dy + w 2 £(8 - y)(2)(y + 6) dy = 2wi£(48 - 14y + y 2 ) dy + 2w 2 £.(48 + 2y - y 2 ) dy 2wi |48y - 7y 2 + \ \ ] + 2w 2 |48y + y 2 - £ | = 216wi + 360w 2 50. (a) F = 62.4/ o 6 (10-y)[(8 I) - (I)] dy 62.4 / (240 - 34y + y 2 ) dy y »6x ^ 240y-17y 2 + ^ =^(1440-612 + 72) 18,7201b. (]fi) (7.6) y » -€x + 48 (7/2,3) • (6,0) (8.0) (7.2) (b) The centroid ( 5 , 3) of the parallelogram is located at the intersection of y = = x and y = — | x -+- 4r . The centroid of 6 v 1 36 5 X -I- 5 . the triangle is located at (7, 2). Therefore, F = (62.4)(7)(36) + (62.4)(8)(6) = (300)(62.4) = 18,720 lb CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES 1. V = 7T Jjf(x)] 2 dx = b 2 - ab => 7r£[f(t)] 2 dt = x 2 - ax for all x > a =^> tt [f(x)] 2 = 2x - a =>• f(x) = ± J^f- 2. V = 7T f [f(x)] 2 dx = a 2 + a =>■ tt f* [f(t)] 2 dt = x 2 + x for all x > a => ?r[f(x)] 2 = 2x + 1 =>- f(x) = ± •J *J 2x+ 1 3. s(x) = Cx => £ v/l + [f'(t)] 2 dt = Cx => a/1 + [f'(x)] 2 = C => f'(x) = ^C 2 - 1 for C > 1 =$► f(x) = f* VC 2 - 1 dt + k. Then f(0) = a => a = + k => f(x) = f ^C 2 - 1 dt + a => f(x) = x^C 2 - 1 + a, where C > 1 . Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Chapter 6 Additional and Advanced Exercises 421 4. (a) The graph of f(x) = sin x traces out a path from (0, 0) to (a, sin a) whose length is L = / \J 1 + cos 2 8 d(9. The line segment from (0, 0) to (a, sin a) has length y/(a — 0) 2 + (sin a — 0) 2 = \J o? + sin 2 a. Since the shortest distance between two points is the length of the straight line segment joining them, we have immediately that J y/l + cos 2 9 dd > \J a 2 + sin 2 a if < a < § . (b) In general, if y = f(x) is continuously differentiable and f(0) = 0, then I y/l + [f'(t)] 2 dt > \Ja 2 + f 2 (a) for a > 0. 5. From the symmetry of y = 1 — x", n even, about the y-axis for — 1 < x < 1 , we have x = 0. To find y M we use the vertical strips technique. The typical strip has center of mass: (x , y ) l-x'M , length: 1 — x", width: dx, area: dA = (1 — x n ) dx, mass: dm = 1 • dA = (1 — x") dx. The moment of the strip about the x-axis is y dm „n>2 1 (n+l)(2n+l)-2(2n + l) + (n+l) €i0 dx = 2| i(T - 2x n + x 2n ) dx = |x- 2n 2 + 3n + I - 4n - 2 + n + 1 _ 2n 2 n+1 x-° + ' 1 1 2n + I J n+1 2n+l (n+l)(2n+l) (n+l)(2n+l) (n+l)(2n+l) Also, M = /_'dA = Jjl - x») dx = 2 X'(l - x») dx = 2 [x - ^] J = 2 (1 n+ 1/ 2n M 2n 2 (n+1) (n+l)(2n+l) 2n 2n+l "^ the limiting position of the centroid is (0, \ o. 2n+l is the location of the centroid. As n n+1 * oo, y Therefore l so 6. Align the telephone pole along the x-axis as shown in the accompanying figure. The slope of the top length of pole is / N.5 _ 9 \ V St i-) 40 8tt 40 (14.5 - 9) 5.5 8?r-40 8ri-S(> . Thus, H) 8- 8,7 y = SF + 8^0 x = t ( 9 + So x ) is an equation of the (*■*) y-^ 9 *^ 80 \ («3r) ■*■ x («-W) line representing the top of the pole. Then, x-^y 2 dx = 7rJ o x[i(9+|ix)] dx = 64-],, ^ (9 + ^ x) dx;M=J ^y 2 dx = " X, 40 [ si ( 9 + 35 x ) ] 2 dx = 5S r" ( 9 + SB x ) 2 dx - Thus ' x = t « fif ~ 23 06 (usin S a caIculator t0 com P ute the integrals). By symmetry about the x-axis, y = so the center of mass is about 23 ft from the top of the pole. (a) Consider a single vertical strip with center of mass ( x , y ). If the plate lies to the right of the line, then the moment of this strip about the line x = b is (x — b) dm = (x — b) 6 dA => the plate's first moment about x = b is the integral J (x - b)<5 dA = J <5x dA - J* <5b dA = M y - b<5A. (b) If the plate lies to the left of the line, the moment of a vertical strip about the line x = b is (b — 3c ) dm = (b — 3c ) S dA => the plate's first moment about x = b is J (b — x)<5 dA = J bS dA — J <5x dA = b<5A - M„. (a) By symmetry of the plate about the x-axis, y = 0. A typical vertical strip has center of mass: (x ,y ) = (x, 0), length: 4^/ax, width: dx, area: 4^/ax dx, mass: dm = S dA = kx - A^fax dx, for some proportionality constant k. The moment of the strip about the y-axis is M y = J 3c dm = / 4kx 2 ^/ax dx = 4k,/aJ" X 5 / 2 Hx = 4k, A n X 7 / 2 ! a = 4 ka l/2 . 2 a 7/2 = 8kal ' x 5 / 2 dx = 4k ^a [f x 7 / 2 ] * = 4ka 2 / 2 • i a 7 / 2 = ^ . Also, M = J dm = J" 4kx ^ax dx = 4k v /^/ o a x 3 / 2 dx = 4k^ [§ x 5 / 2 ] J = 4ka J / 2 . § a 5 / 2 = ^ . Thus, x => (x, y) = (y , 0) is the center of mass. (b) A typical horizontal strip has center of mass: (x ,y ) = I 4a 2 a , y J = ( 8ka 4 Ska ' y 2 +4a 2 8a ' y ) , length: a - fj width: dy, area: ( a — J- J dy, mass: dm = 6 dA = |y | (a — |- J dy. Thus, M x = J y dm Copffigl (c) 1 Pearson Etation, Inc., publishing as Pearson Addison-Wesle 422 Chapter 6 Applications of Definite Integrals a „3 r 20a M £ y ly I (a - i) dy = £ a -y 2 (a - g) dy + £ y 2 (a - g) dy /l(-ay 2 + ^)dy + /;(ay 2 -^)dy=[ -¥ + ^ + ¥"S = 0:M, = /*dm=£;(^!)|y|(a-g)dy i / Jy| (y 2 + 4a 2 ) (*^) dy = ^ / |y| (16a 4 - y 4 ) dy V ' -2a 3^£ a (-16a 4 y + y 5 )dy+3^X 2a (16a 4 jfc [8a 4 - 4a 2 - if] + £ [ 8 a 4 - 4a 2 - if] = £ ( 3 2a' = / dm = iIiy|(^) d y = siIiy|( 4a2 -y 2 ) d y s l_J- 4a2 y + y 3 ) d y + k C( 4a2 y - y 3 ) d y = k [- 2a 2 • i ^2a 2 • 4a 2 - ±f-*) = i (8a 4 - 4a 4 ) = 2a 3 . Therefore, x 1 y-y 5 )dy=3^|-8a 4 y 2 ~>- + ^P 8-V-J ,6 _ 32a 6 ,2 y 2 + £ -2a sP ' 3 (32a ) = 3 a ; !2a 2 y 2 -^ M )(£) °f and y = ^ = is the center of mass. 9. (a) On [0, a] a typical vertical strip has center of mass: (x , 'y ) = ( x, — — ~ x ^ a ~ x j , length: y b 2 — x 2 — ya 2 — x 2 , width: dx, area: dA = I y b 2 — x 2 — y a 2 — x 2 J dx, mass: dm = 6 dA = 6 I v b 2 — x 2 — v a 2 — x 2 J dx. On [a, b] a typical vertical strip has center of mass: (x , y ) = (x, ^ mass: length: y b 2 — x 2 , width: dx, area: dA = y b 2 — x 2 dx, dm = 6 dA = S ^b 2 - x 2 dx. Thus, M x = / y dm £\ ( v / b 2 -x 2 + Va 2 - x 2 ) 6 (Vb 2 -x 2 - a/ a 2 - x 2 ) dx + _£" 1 x/b 2 - x 2 6 ^b 2 - x 2 dx Jj(b 2 - x 2 ) - (a 2 - x 2 )] dx + | J* (b 2 - x 2 ) dx = | Jjb 2 - a 2 ) dx + § J* (b 2 - x 2 ) dx [(b 2 - a 2 ) x] a + f [b 2 x - f ] " = | [(b 2 - a 2 ) a] + § [(b 3 - f ) - (b 2 a - f )] (ab 2 - a 3 ) + f (§ b 3 - ab 2 + f ) = ^ - f = «5 (^) ; M y = / x dm J\o (yjb 2 -x 2 - A/a 2 - x 2 ] dx + f\d s/b 2 - x 2 dx * f 3 x (b 2 - x 2 ) 1/2 dx - 6 P x (a 2 - x 2 ) 1/2 dx + 6 f x (b 2 - x 2 ) 1/2 dx «-/ <-/ *-* it 2(a 2 -x 2 ) 3 - 2 (b 2 - x 2 ) 3 2 3 J n 2 [b 2 - x 2 ) 3 2 3 „2\ 3 / 2 Iu3\ 3 / 2 .2*3/2 (b 2 -a 2 ) 0/ -(b 2 )°^ +f 0-(a 2 ) We calculate the mass geometrically: M = 6 A = S I 0-(b 2 -a 2 ) 2x3/2 fibi _ fa! _ Mb 3 -a 3 ) 3 3 — 3 M x ; -b 2 ( 7ra 2 \ 6tt < 4 ) ~ 4 v 6 ^ =^(b 2 -a 2 ). Thus,x=^ _ a (b 3 - a 3 ) 4 A ( \?-£ \ ~ 3 ' <57r(b 2 -a 2 ) ~~ 3tt ^ b 2 - a 2 ^ tt _ Mx _ 4 (a 2 +ab+b 2 ) " M 3jr(a+b) n2 i .,2 i .,2 4 (b - a) (a 2 + ab + b 2 ) _ 4 (a 2 + ab + b 2 ) 37T (b - a)(b + a) 3?r(a + b) ; likewise (b) b h T a £ ( 3 ^T^) = (s) (***) = (£) (f) = f =* WW = (*,*) - *e limiting position of the centroid as b — > a. This is the centroid of a circle of radius a (and we note the two circles coincide when b = a). Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle Chapter 6 Additional and Advanced Exercises 423 10. Since the area of the traingle is 36, the diagram may be labeled as shown at the right. The centroid of the triangle is (f , y). The shaded portion is 144 - 36 = 108. Write (x, y) for the centroid of the remaining region. The centroid of the whole square is obviously (6, 6). Think of the square as a sheet of uniform density, so that the centroid of the square is the average of the centroids of the two regions, weighted by area: 36(f) + 108{x) _„ Je 36(f) +108(y) 6 and 6 144 """ " — 144 which we solve to get x = 8 — | and y = — x = 7 in. (Given). It follows that a = 9, whence y Set 04 9 7 £ in. The distances of the centroid (x, y) from the other sides are easily computed. (Note that if we set y = 7 in. above, we will find x 7 1 ) ' 9-> 11. 2v/x" ds ldx JTv^ ldx= | [(1+x) 3 / 2 ] 28 3 12. This surface is a triangle having a base of 27ra and a height of 27rak. Therefore the surface area is \ (27ra)(27rak) = 27r 2 a 2 k. 2 d£x dt- 13. F = ma = t x = when t = n(\l W=fFdx = J o _ (12mh) 3 / 2 a => Ci (^mh) 1 /* 12mh- v 12mh V X = dx dt F(t) - £ dt : Jo 12m (12mh)V4 3 m Then x C; v = when t = => C = =*> dx dt 3 m h =*► t 2 dt (12mh) 1//4 . The work done is (12mh)V4 /ix r/, = Gib (12mh)«/ 4 18n 18m f -2^3mh V3mh t* 12m •Ci; 14. Converting to pounds and feet, 2 lb/in 21b 12 in 1 in 1 ft 24 lb/ft. Thus, F = 24x W n 1/2 / 24x dx J() fl2x 21 1/2 Jo 3 ft • lb. Since W = \ mvg - \ mvf , where W = 3 ft • lb, m = l-k lb v 32 ft/sec 2 / = 35q slugs, and Vi = ft/sec, we have 3 s = - 16t 2 + v t (since s = at t = 0) => ,32^ and the height is s = — 16 ( — vo(jf) = (&) ~ 64 'J- v 2 ) ,320 *0) 3-640 64 Vg = 3 • 640. For the projectile height, 32t + v . At the top of the ball's path, v = = 30 ft. hi 32 15. The submerged triangular plate is depicted in the figure at the right. The hypotenuse of the triangle has slope — 1 => y — (—2) = — (x — 0) => x = — (y + 2) is an equation of the hypotenuse. Using a typical horizontal strip, the fluid pressure isF = /(62.4).( ( gP i ).( 1 ^)dy = J_" 6 2 (62.4)(-y)[-(y + 2)] dy = 62.4 f~*(y 2 + 2y) dy 62.4 | C + y 2 (62.4) ' 208 -6 32) ( 62.4)[(-f+4)-(-f +36)] (62.4X112) 2329.6 lb x--(y + 2) (4,-6) Copyright (c) 2006 Pearson Education 424 Chapter 6 Applications of Definite Integrals 16. Consider a rectangular plate of length I and width w. The length is parallel with the surface of the fluid of weight density u>. The force on one side of the plate is F = ujJ^ (-y)(£) dy = -U [£] ' = ^f- . The average force on one side of the plate is F ilv = ^ J (— y)dy if . Therefore the force ^f- -y -w (£w) = (the average pressure up and down) • (the area of the plate). one side 17. (a) We establish a coordinate system as shown. A typical horizontal strip has: center of pressure: (x ."y ) = (|,y) , length: L(y) = b, width: dy, area: dA = b dy, pressure: dp — uj |y| dA = uib |y| dy => F x = J y dp = J_ h y • ub |y| dy = -wb J y 2 dy -ujb -ub -h 3 3 -ubh 3 . 3 F = J dp = J_ h u |y| L(y) dy = -wb J_ h y dy b,-h) -o;b -wb 0- h 2 ^f . Thus, y = | 2 J (b) A typical horizontal strip has length L(y). By similar triangles from the figure at the right, -4p = ~ y h ~ a L(y) jr (y + a). Thus, a typical strip has center of pressure: (x ,y ) = (x , y), length: L(y) = — j(y + a), width: dy, area: dA = - | (y + a) dy, pressure: dp = uj |y| dA = uj{— y) (— ^) (y + a) dy f (y 2 + ay) dy => F x = / y dp Ll w y f (y 2 + a y) d y = f J ,.,... (y :i + u V -ui> -(a+h) ' h 4 ' 3 ^'b P (y 3 J -(a+h) W -(a+h) HO^H 1 a + h) 4 a(a + h) 3 )] uM | a 4 -(a + h) 4 h -2h 3 -a(a + h) 3 the distance below the surface is | h. (b,-a-h) 12h ^'b 12h [3 (a 4 - (a 4 + 4a 3 h + 6a 2 h 2 + 4ah 3 + h 4 )) - 4 (a 4 - a (a 3 + 3a 2 h + 3ah 2 + h 3 ))] (12a 3 h + 12a 2 h 2 + 4ah 3 - 12a 3 h - 18a 2 h 2 - 12ah 3 - 3h 4 ) = f| (-6a 2 h 2 - 8ah 3 - 3h 4 -ff (6a 2 + 8ah + 3h 2 ) ; F = / dp = / u |y | L(y) dy = f £^ h) (y 2 + ay) dy = f [£ + *£] i±* \( ^t. _|_ S 3 .^ _ f -(a + h) 3 , a(a + h) 2 Nl _ yb [~ (a + h) 3 -a 3 _,_ a 3 -a(a + h) 2 ] u* |" a 3 + 3a 2 h + 3ah 2 + h 3 -a 3 , a 3 - (a 3 + 2a 2 h + ah 2 ) h [ 3 + 2 ^ (6a 2 h + 6ah 2 + 2h 3 - 6a 2 h - 3ah 2 ) = ^ (3ah 2 + 2h 3 ) = ^ (3a + 2h). Thus, y = | ■(a+h) 3^2 ^ [2 (3a 2 h + 3ah 2 + h 3 ) - 3 (2a 2 h + ah 2 )] =^)(6a 2 + 8ah + 3h 2 ) _ , _y (^)(3a + 2h) ~~ V"2". ( 6a 2 + 8ah + 3h 2 \ V_ 3a + 2h J the distance below the surface is 6a 2 + 8ah + 3h 2 6a + 4h Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle CHAPTER 7 TRANSCENDENTAL FUNCTIONS 7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES 1 . Yes one-to-one, the graph passes the horizontal test. 2. Not one-to-one, the graph fails the horizontal test. 3. Not one-to-one since (for example) the horizontal line y = 2 intersects the graph twice. 4. Not one-to-one, the graph fails the horizontal test. 5. Yes one-to-one, the graph passes the horizontal test 6. Yes one-to-one, the graph passes the horizontal test 7. Domain: < x < 1, Range: < y 8. Domain: x < 1, Range: y > y y-f(x) 9. Domain: -1 < x < 1, Range: - § < y < 10. Domain: — oo < x < oo, Range: y y = f(x)/ -f<y<l 11. The graph is symmetric about y = x. A V ~""\ y=JT7 \ 0<x<l 1 (b) y = \/\ -x 2 => y 2 = 1 - x 2 => x 2 = 1 - y 2 => x = y/l - y 2 => y = v 7 ! - x 2 = f _1 (x) Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 426 Chapter 7 Transcendental Functions 12. The graph is symmetric about y = x. x = \ => y = i = f -1 tt 13. Step 1: y = x 2 +l=^x 2 = y-l^x = y/y - 1 Step 2: y = Vx - 1 = f _1 (x) 14. Step 1: y = x 2 => x = — a/y, since x < 0. Step 2: y = - v /^=f- 1 (x) 15. Step 1: y Step 2: y x 3 - 1 =>■ x 3 = y ■ VxTI = f-^x) (y+ 1)1/3 16. Step 1: y = x 2 - 2x + 1 =>• y = (x - l) 2 Step 2: y = 1 + yfx = f^x) =4- y/y = x — 1, since x > 1 =>- x=l + -/y 17. Step 1: y = (x + l) 2 =4> ^/y = x+ 1, since x > -1 =>• x = y^y- 1 Step 2: y= y^-l = f- 1 (x) 18. Step 1: y = x 2 / 3 => x = y 3 / 2 Step 2: y = x 3 / 2 = f- : (x) 19. Step 1: y = x 5 => x = y 1 / 5 Step 2: y = \/x = f ^(x); Domain and Range of f _1 : all reals; f(f-!(x)) = (x 1 / 5 ) 5 = x and f -1 (f(x)) = (x 5 ) 1/5 = x 20. Step 1: y = x 4 x = y 1/4 Step 2: y = S/x = f" 1 ^); Domain of f _1 : x > 0, Range of f -1 : y > 0; f(f-!(x)) = (x 1 / 4 ) 4 = xandf-^fW) = (x 4 ) 1/4 = x 1/3 21. Step 1: y = x 3 + 1 =>• x 3 = y - 1 => x = (y - 1) Step 2: y = V* - 1 = f _1 (x); Domain and Range of f _1 : all reals; f(f-!(x)) = ((x- l) 1 / 3 ) 3 + 1 =(x- 1)+ 1 =xandf- 1 (f(x))= ((x 3 + 1) - 1) 1/3 = (x 3 ) 1/3 = x Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 7.1 Inverse Functions and Their Derivatives 427 22. Stepl: y = |x — | =>• |x = y+| =4> x = 2y + 7 Step 2: y = 2x + 7 = f -1 (x); Domain and Range of f _1 : all reals; f(f-i(x)) = I(2x + 7)-| = (x+Z) -| = xandf- 1 (f(x)) = 2(ix- |) + 7 = (x - 7) + 7 = x 23. Step 1: y Step 2: y l f- J (x) Domain off 1 : x > 0, Range off 1 : y > 0; f(f- 1 (x)) = y^ = ^ = xandf-^fW) - ' ,7^) x since x > 24. Stepl: y = 4j^x 3 =i^x=^3 Step 2: y= J 7 3 = yi = f-i(x); Domain of f _1 : x^O, Range of f _1 : y ^ 0; f(f- 1 (x)) (x-l/3)' 1 i = x and f- J (f(x)) -1/3 25. (a) y = 2x + 3 => 2x = y - 3 => X = | - | =* f-!(x) =|-| df | _ 2 df- 1 | ._ 1 (C) dx I x=— 1 dx (b) ^y=/(x) = 2x + 3 26. (a) y = \ x + 7 =4> y-7 (c) => x = 5y - 35 => f-\x) = 5x - 35 df I _ 1 df^ I _ g dxL=-i 5' dx | x=34/5 (b) 27. (a) y = 5 - 4x =>■ 4x = 5 - y f-!(x) = 5 _ y 4 4 5 _ i 4 4 (C) df I dx I x =l/2 H ' dx (b) Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle 428 Chapter 7 Transcendental Functions 28. (a) y = 2x 2 9 1 x = y (c) — I dx I x=5 df-' l dx y 2 ^ => f-^x) = 20, /2 4x -4= X-V2 2\/2 J_ 20 (b) 0.5 1 29. (a) f(g(x)) = (\/x") 3 = x, g(f(x)) = Vx» = x (c) f (x) = 3x 2 => f (1) = 3, f (-1) = 3; g'(x)=ix- 2 / 3 => g'(l)=I, g '(-l) = I (d) The line y = is tangent to f(x) = x 3 at (0, 0); the line x = is tangent to g(x) = 3 x/x at (0, 0) (b) 30. (a) h(k(x))=i((4x) 1 /3) 3 = x , k(h(x)) = (4 • fj 1/3 (c) h'(x) is 2 h'(2) = 3, h'(-2) = 3; k'(x) = I (4x)- 2 / 3 => k'(2) = 1, k'(-2) = i (d) The line y = is tangent to h(x) = j at (0, 0); the line x = is tangent to k(x) = (4x) 1,/3 at (0,0) (b) 3 X J 1"T y = (4x) 1/3 31. df 3x 2 — 6x df- 1 dx _ J_ — df = f(3) dx 32. df 2x-4 => df 1 dx 33. df- df- 1 dx < = f(2) dx J- - 3 (I) " 3 34. dg- 1 I dg- 1 dx _ J_ — dg lx = f(0) 37 35. (a) y = mx =>• x = ± y => f x (x) = 5 x (b) The graph of y = f _1 (x) is a line through the origin with slope — . 36. y = mx + b =>- x = — — — =>• f _1 (x) = — x — -; the graph of f _1 (x) is a line with slope — and y-intercept 37. (a) y = x + 1 =>• x = y - 1 =>• f- a (x) = x - 1 (b) y = x + b =>• x = y-b => f _1 (x) = x - b (c) Their graphs will be parallel to one another and lie on opposite sides of the line y = x equidistant from that line. 'y = x Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.1 Inverse Functions and Their Derivatives 429 38. (a) y = -x+l => x = -y + 1 => f-\x) = 1 - x; the lines intersect at a right angle (b) y = -x + b => x=-y + b =>- f _1 (x) = b - x; the lines intersect at a right angle (c) Such a function is its own inverse. 39. Let Xi ^ x 2 be two numbers in the domain of an increasing function f. Then, either Xj < x 2 or xi > X2 which implies f(xi) < f(x 2 ) or f(xi) > f(x 2 ), since f(x) is increasing. In either case, f( x i) 7^ f( x 2) and f is one-to-one. Similar arguments hold if f is decreasing. 40. f(x) is increasing since x 2 > Xi I x „ + 5 > l,5.df_l 3 x 2 "I" 6 -? 3 X! -|- g , dx — 3 df" 1 _ J_ dx " (i) 41. f(x) is increasing since x 2 > x : =>• 27X 3 , > 27xJ; y = 27x 3 => x = ± y 1 / 3 => f : (x) = | x 1/3 ; £ = 8 lx2 df 1 I _ _J_ _ 1 y -2/3 dx "~ Six 2 I i x V3 ~~ 9x 2 / 3 — 9 A 42. f(x) is decreasing since x 2 > Xj =^> 1 — 8xf < 1 - 8xJ; y = 1 - 8x 3 => x = \ (1 - y) 1/3 =>■ f ^x) = I (1 - x) 1/3 ; df dx -24x 2 df- 1 1 -1 dx " -24x 2 U(i-x)V3 6(1 -x) 2 / 3 i(l-x)- 2 / 3 43. f(x) is decreasing since x 2 > Xj =4> (1 - x 2 ) 3 < (1 - Xi) 3 ; y = (1 - x) 3 =^ x = 1 - y 1 / 3 => f : (x) = 1 ,1/3. -3(1 - x) 2 df 1 dx -3(1 -x) 2 -1 3x 2 3 l x -2/3 44. f(x) is increasing since x 2 > Xi => x 2 ' > x/ ; y = x 5 / 3 =>■ x = y 3 / 5 =>• f : (x) = x 3 / 5 ; df _ 5 Y 2/3 dx — 3 A df 1 dx 1 \ X 2 / 3 3 _ 3 y-2/5 5x 2/5 - 5 x 45. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if Xi ^ x 2 then f( x i) 7^ f( x 2)> so — f( x i) 7^ — f( x 2) an d therefore g(xj) ^ g(x 2 ). Therefore g(x) is one-to-one as well. 46. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if Xi ^ x 2 then f(xO ^ f(x 2 ), so ^ ± ^ , and therefore h(xi) ^ h(x 2 ). 47. The composite is one-to-one also. The reasoning: If Xi ^ x 2 then g(xi) ^ g(x 2 ) because g is one-to-one. Since g( x i) 7^ g( x 2), we also have f(g(xi)) ^ f(g(x 2 )) because f is one-to-one. Thus, f o g is one-to-one because Xi ^ x 2 =* f(g( Xl )) / f(g(x 2 )). 48. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers Xi ^ x 2 in the domain of g with g(xi) = g(x 2 ). For these numbers we would also have f(g(xi)) = f(g(x 2 )), contradicting the assumption that f o g is one-to-one. Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 430 Chapter 7 Transcendental Functions 49. The first integral is the area between f(x) and the x-axis over a < x < b. The second integral is the area between f(x) and the y-axis for f(a) < y < f(b). The sum of the integrals is the area of the larger rectangle with corners at (0, 0), (b, 0), (b, f(b)) and (0, f(b)) minus the area of the smaller rectangle with vertices at (0, 0), (a, 0), (a, f(a)) and (0, f(a)). That is, the sum of the integrals is bf(b) — af(a). f(b) 50. f'(x) = ^ cx + 1 — v*2 + = f^ — ^j. Thus if ad — be ^ 0, f'(x) is either always positive or always negative. Hence f(x) is either always increasing or always decreasing. If follows that f(x) is one-to-one if ad — be ^ 0. 51. (go f)(x) = x => g(f(x)) = x => g'(f(x))f'(x) = 1 52. W(a) = f\ [(f^Hy)) 2 - a 2 ] dy = = £ 2?rx[f(a) - f(x)] dx = S(a); W'(t) = ^[(f-^fCt))) 2 - a 2 ] f'(t) = 7T (t 2 - a 2 ) f '(t); also S(t) = 27rf(t)/ a ' x dx - 27r/' xf(x) dx = [7rf(t)t 2 - 7rf(t)a 2 ] - 27r/' xf(x) dx => S'(t) = 7rt 2 f'(t) + 27rtf(t) - 7ra 2 f'(t) - 27rtf(t) = 7T (t 2 - a 2 ) f'(t) => W'(t) = S'(t). Therefore, W(t) = S(t) for all t G [a,b]. 53-60. Example CAS commands: Maple : with( plots );#53 f := x -> sqrt(3*x-2); domain := 2/3 ..4; xO := 3; Df := D(f); # (a) plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[l,3], legend=["y=f(x)","y=f '(x)"], title="#53(a) (Section 7.1)" ); ql := solve( y=f(x), x ); # (b) g:=unapply(ql,y ); ml := Df(x0); # (c) tl :=f(x0)+ml*(x-x0); y=tl; m2 := 1/Df(x0); # (d) t2 := g(f(x0)) + m2*(x-f(x0)); y=t2; domaing := map(f, domain); # (e) pi := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[l,9], thickness=[3,0] ): p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): p3 := plot( tl, x=x0-l..x0+l, color=red, linestyle=4, thickness=0 ): p4 := plot( t2, x=f(x0)-l..f(x0)+l, color=blue, linestyle=7, thickness=l ): p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ): display( [pl,p2,p3,p4,p5], scaling=constrained, title="#53(e) (Section 7.1)" ); Mathematica: (assigned function and values for a, b, and xO may vary) If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica to do this. See section 2.5 for details. «Miscellaneous "RealOnly" Clear[x, y] Copyright (c| 1 Pearson E Won, Inc., publishing as Pearson Addison-Wesle Section 7.1 Inverse Functions and Their Derivatives 43 1 {a,b} = {-2, 1};x0= 1/2; f[x_] = (3x + 2) / (2x - 11) Plot[{f[x],f[x]}, {x, a,b}] solx = Solve[y == f[x], x] g[yj = x/. solx[[l]] yO = f[xO] ftan[x_] = yO + f [xO] (x-xO) gtan[y_] = xO + 1/ f [xO] (y - yO) Plot[{f[x], ftan[x], g[x], gtan[x], Identity [x]},{x, a, b}, Epilog -+ Line[{{xO, yO},{yO, xO}}], PlotRange -> {{a,b},{a,b}}, AspectRatio Automatic] 61-62. Example CAS commands: Maple : with( plots ); eq := cos(y) = x A (l/5); domain := .. 1; xO := 1/2; f := unapply( solve( eq, y ), x ); # (a) Df := D(f); plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[l,3], legend=["y=f(x)","y=f (x)"], title="#62(a) (Section 7.1)" ); ql := solve( eq, x ); # (b) g:=unapply(ql,y ); ml := Df(xO); # (c) tl :=f(xO)+ml*(x-xO); y=tl; m2 := 1/Df(x0); # (d) t2 := g(f(xO)) + m2*(x-f(x0)); Y=t2; domaing := map(f, domain); # (e) pi := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[l,9], thickness=[3,0] ): p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): p3 := plot( tl, x=xO-l..xO+l, color=red, linestyle=4, thickness=0 ): p4 := plot( t2, x=f(xO)-l..f(xO)+l, color=blue, linestyle=7, thickness=l ): p5 := plot( [ [xO,f(xO)], [f(xO),xO] ], color=green ): display( [pl,p2,p3,p4,p5], scaling=constrained, title="#62(e) (Section 7.1)" ); Mathematica: (assigned function and values for a, b, and xO may vary) For problems 61 and 62, the code is just slightly altered. At times, different "parts" of solutions need to be used, as in the definitions of f[x] and g[y] Clearfx, y] {a,b) = {0, 1};x0=1/2; eqn = Cos[y] == x 1/5 soly = Solve[eqn, y] f[x_] = y /. soly[[2]] Plot[{f[x], f[x]}, {x, a, b}] solx = Solve[eqn, x] g[yj = x/. solx[[l]] yO = f[xO] ftan[x_] = yO + f [xO] (x - xO) Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 432 Chapter 7 Transcendental Functions gtan[y_] = xO + 1/ f [xO] (y - yO) Plot[{f[x], ftan[x], g[x], gtan[x], Identity [x]},{x, a, b}, Epilog -» Line[{{xO, yO},{yO, xO}}], PlotRange -» {{a, b}, {a, b}}, AspectRatio ->• Automatic] 7.2 NATURAL LOGARITHMS 1. (a) In 0.75 = In \ = In 3 - In 4 = In 3 - In 2 2 = In 3 - 2 In 2 (b) In | = In 4 - In 9 = In 2 2 - In 3 2 = 2 In 2 - 2 In 3 (c) In \ = In 1 - In 2 = - In 2 (d) In \/9 = \ In 9 (e) In 3v/2 = In 3 + In 2 1 / 2 = In 3 + \ In 2 (f) In ^/Y15 = \ In 13.5 = ± In f = \ (In 3 3 - In 2) = ± (3 In 3 - In 2) 2. (a) In jij = In 1 - 3 In 5 = -3 In 5 (c) In lyfl = In 7 3 / 2 = | In 7 (e) In 0.056 = In ^ = In 7 - In 5 3 = In 7 - 3 In 5 In 35 + In 1 I In 3 2 = | In 3 (b) In 9.8 = In f = In 7 2 - In 5 = 2 In 7 - In 5 (d) In 1225 = In 35 2 = 2 In 35 = 2 In 5 + 2 In 7 (f) In 25 In 5 + In 7 - In 7 _ 1 2 In 5 2 3. (a) In sin 9 — In ' sin 8 \ In sin In 5 (b) ln(3x 2 -9x)+ln(i) = In (^*s) =ln(x-3) (c) i In (4t 4 ) - In 2 = In v^ - In 2 = In 2t 2 - In 2 = In (?f\ = In (t 2 ) 4. (a) In sec + In cos 9 = In [(sec 6>)(cos 9)] = In 1 = (b) In (8x + 4) - In 2 2 = In (8x + 4) - In 4 = In (^f 1 ) = In (2x + 1) (c) 3 In Vt 2 - 1 - In (t + 1) = 3 In (t 2 - 1) 1/3 - In (t + 1) = 3 (±) In (t 2 - 1) - In (t + 1) = In ( (t + ( ,'f ,7 1} ) = ln(t-l) 5. y = ln3x => y'=(i)(3)=i 7. y = ln(t 2 ) => |=(i)(2t) = f 9. y = ln l = i n3x -i => | = (j^) (-3x- 2 ) = - 1 10. y = 1b 12 = In 10x-l =*£=(,£*) (-10x- 2 ) = - 1 11. y = ki(0 + l) => I = (yi-f) (1) = ^ 12. y = ln(20 + 2) 6. y = lnkx => y'= (i)(k) = x 8 . y = in(t 3 / 2 ) => % = (&)am = i dt VtW \2 dy = dfl V 29 + 2 x,)(2) ' + 1 13. y = lnx 3 => I = (i) (3x 2 ) = | 14. y = (in x) 3 =* | = 3(ln x) 2 - £ (In x) = ^ Jy 15. y = t(ln t) 2 =^ f = (In t) 2 + 2t(ln t) • ^ (In t) = (In t) 2 + ^fi = (In t) 2 + 2 In t 16. y = tv/tat = t(ln t) 1 / 2 =» g = (In t) 1 / 2 + I t(ln t)' 1 / 2 - A (In t) = (In t) 1 ' 2 + !2e|L^ = (In t) 1 / 2 2(ln t)V2 17. y = ^lnx-^ =* S =x 3 lnx+ x 1 4x 3 4 x 16 x 3 In x Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.2 Natural Logarithms 433 18. y = f In x - f => | = x 2 In x + ^ • i - ^ - v- x J In x iq v _ lnt . dy _ t(|)-(lnt)(l) _ t _ ln , on x, - 1 + lnt . dy _ t(j)-(l + lnt)(l) _ i _ i _ ln t _ tot ZU. y - ( =x> dt - (2 - (2 - ,2 O, v _ Inx ^ .,,_ (l+lnx)(l)-(lnx)(l) 1 + '^-^ Z ' i - y l+lnx ^ > (1+lnx) 2 (1+lnx) 2 x(l+lnx) 2 09 „ - xlnx . ,J _ (l+lnx)(lnx + x-l)-(xlnx)(l) _ (l+lnx) 2 -lnx _ , lnx ZZ - y — 1+lnx ^ y - ,.,.-_« - 7i ,.-..« - I (1+lnx) 2 (1+lnx) 2 (1+lnx) 2 23. y = ln(lnx) =>• y' « In x/\x7 xlnx 24. y = ln(ln(lnx)) => y ' — 1 . _d_ ln (ln x) dx (In (lnx)) 1 1 d ln (ln x) ln x dx (lnx) l x (ln x) ln (ln x) 25. y = #[sin (ln 9) + cos (ln &)]=>% = [sin (ln 9) + cos (ln 9)] + 6 [cos (ln 0)-\- sin (ln 9) ■ \] = sin (ln 9) + cos (ln 9) + cos (ln 9) - sin (ln 9) = 2 cos (ln 8) 26. y = ln (sec 9 + tan 6) dy _ sec 8 see 8 tan 6 + sec z sec 8 + tan sec 6>(tan + sec f tan 8 + sec # sec 8 27. y xi/s+ 1 lnx- i ln(x+ 1) 1 / l * 2 \\+\, 2(x+l)-x 2x(x+ 1) 3x + 2 2x(x+l) 28. y=ilni^ = I[ln(l+x)-ln(l-x ) ] =» y' = \ [^ - (^) (-1)] = \ [fa -x+l+x 1 x)(l - x) 1 - X 2 29. y 1+lnt . dy (l-lnt)(j) -(l+lnt)(=l) 1-lnt ^ dt (1-lnt) 2 1 In t i 1 i In t t 1 ' t ^ ' (1-lnt) 2 t(l-lnt) 2 30. y = Jin y/x = (in t V 2 ) 1/2 =* | = \ (in tV 2 ) ^ ■ | (in t V 2 ) = i (in t^ 2 ) ^ 2 - £ - - (t^) In t l/2\" 1 / 2 _ J^_ _ 1 -1/2 _ 1_ tl/2 2 4tv/ln y't 31. y = ln (sec (ln 8)) % = -^n» ■ f H (sec(ln 8)) = sec (ln 9) n ta " (ln g) • A (ln 0) = ^ d<9 sec (In 0) d# v v 7/ sec(ln#) 66 v 7 32. y = ln ^^"g 9 = § (ln sin 9 + ln cos 8) - ln (1 + 2 ln 0) dy 1 / cos g _ sin g \ _ 9 &B ~ 2 V sin 8 cos 9 J 1 + 2 ln ( cot 8 — tan i 4 (l + 21n£ 33. y = ln (x 2 +l)-' (/I -X 51n(x 2 + l)-iln(l-x) =* y' = ^ - \ (^) (-1) = jfc + ^ "■ v-ln A /^ = |[5 1n(x+l)-201n(x + 2)]^y' = l(^T-xf2) = I fefl^ 5 3x + 2 2 (x+l)(x + 2) 35 ' ? = £„ '" <A«K => I = (m v^) ■ i (x 2 ) - (ln Jf) ■ £ (f ) = 2x In |x| - x In $ Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 434 Chapter 7 Transcendental Functions 36. y = ff hit* =► | = (in \A) • £ (\^) - (in ^) • & (^ = (in \A) (|^ 3 ) - (in ^) (^ /^ In f/x In \/x 3^2 2^ 37. J* * i dx = [In |x|] Ig = In 2 - In 3 = In | 39 - J ? ^dy = ln|y 2 -25|+C 38. / ^ 5^2 dx = [In [3x - 2|] ° 1 = In 2 - In 5 = In § 40 - /4^5dr = ln|4r 2 -5|+C 41. /''j^h dt = [l n I 2 ~~ cos t[] o = In 3 - In 1 = In 3; or let u = 2 - cos t => du = sin t dt with t = =S> u = 1 and t = 7T => u = 3 =>• /" ^^ dt = J] 1 du = [ln |u|] \ = ln 3 - ln 1 = ln 3 42. P \ttlie d0= M 1 -4cos6»|]o /3 = In [1 - 2| = -ln 3 = ln \\ or let u = 1 - 4 cos 9 => du = 4 sin 9 d<9 with 61 = =>■ u = -3 and 9 = § =>• u = - 1 =>■ J" * y^f^g d0 = /_"' 1 du = [ln |u|] I* = - ln 3 = ln 1 43. Let u = ln x => du = ~ dx; x = 1 =>■ u = and x = 2 =4> u = ln 2; £l_X^ dx = rj" 2 2u du = [ U 2]M = (ln 2) 2 44. Let u = ln x =>• du = ~ dx; x = 2 =>• u = ln 2 and x = 4 => u = ln 4; £ n^ = IT u du = [ ln «] " = ln ( ln 4) - In (ln 2) = ln (HI) = ln (^f ) 45. Let u = ln x => du = - dx; x = 2 =>• u = ln 2 and x = 4 => u = ln 4; ln(^)=ln2 J» 4 n ln 4 , * = f u- 2 du = - i V 2 x(lnx) 2 J ln2 L uJ ] 1 , j_ _j_ 1 j_ 1 1 1 1 In 4 ~r ln 2 ~~ ln 2 2 ~*~ ln 2 — 2 In 2 + ln 2 — 2 ln 2 — ln 4 46. Let u = ln x => du = - dx; x = 2 => u = ln2 and x = 16 => u = ln 16; /; dx 2x\/ln x 2 f " '" u- 1 / 2 du = [u 1 / 2 ! '" '' = x/ln 16 - V/LT2 = x/4 In 2 - v/hT^ = 2v/ln2 - \f\nl = v/In2 J 1„ 2 L J ln2 v v v v v v v 47. Let u = 6 + 3 tan t =>• du = 3 sec 2 1 dt; /6^ dt =/T=Hu| + C = ln|6 + 3tant|+C 48. Let u = 2 + sec y =>• du = sec y tan y dy; |f^ d y = /T=l n H+C = ln|2 + secy|+C 49. Let u = cos | => du = - J sin | dx => -2 du = sin £ dx; x = => u = 1 and x = 5 => u = -4- 2 2 2 2 2 yj2 -Wft, tan f dx = I - — \ dx 2 J() cos 2 2 jC^T = h 2 ln M \'^ = ~ 2 ln 775 = 2 1° V 72 = l n 2 p7T/2 /»: I tan | dx = I Jo 2 Jo 50. Let u = sin t =4> du = cos t dt; t = f =>• u = -4= and t = ? =>■ u = 1 ; 4 , /? 2 J'" - / 2 r*V 2 r 1 1 cottdt= / £?^dt= / ,-^ = [ln|u|] ir/4 Jtt/1 slnt J U\fi u ! -ln-L=ln^2 51. Let u = sin | =>■ du = i cos f d0 =^ 6 du = 2 cos | d0; 9 = f => u = \ and = tt => u = ^ ; X> cot J d ^ = X/2 ? £r de = 6 / lA ^=6[ln|u|]^ = 6(lnf -lnl)=61nv^ = ln27 Copy |t (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.2 Natural Logarithms 435 52. Let u = cos 3x =>• du = —3 sin 3x dx => —2 du = 6 sin 3x dx; x = =>• u = 1 and x = -^ => u 12 /2' I 6 tan 3x dx = | Jo Jo 6 sin 3x cos 3x dx -'/ du u -2[ln|u|] 2 In 4- - In 1 = 2 In V 2 = In 2 72 v 53. f dx = r J 2\/x + 2x J dx ; let u = 1 + ,/x ^ du = -V dx; f dx 2VTpTVJ) In |1 + Jx\ + C = In (1 + Jx) + C 2v/x"(l + v^) /&=ln|u| + C 54. Let u = sec x + tan x => du = (sec x tan x + sec 2 x) dx = (sec x)(tan x + sec x) dx =>- sec x dx = — ; secxdx . f_^_ = r (ln u) -i/2 . I du = 2 (ln u) 1 / 2 + C = 2^/ln(sec x + tan x) + C J uvlnu J u v Ay/ln (sec x + tan x) / 55. y = v/x(xTT) = (x(x + l)) 1 / 2 => lny = \ ln(x(x+ 1)) => 2 In y = ln(x) + ln(x + 1) =^ v' — ( 1 \ /vCv 4_ n ( 1 i M — V / "( x + 1 )( 2x + 1 ) _ 2x + l =► y - UJ V X ( X + 1) U + x+lJ - S(xTTj - 2Vx(x+l) X ' x+1 56. y= VU 2 + 1)( X -D 2 => lny=i[ln(x 2 +l) + 21n(x-l)] =* L = 1 (^ + ^ y' = v/(x 2 + i) (x - i) 2 (^ + ^) = V(* 2 + i) (x - i) 2 [fenf^u] = ^vSt^ (2x 2 -x+l) x — 1 1) 57. y L_-f_t_^ ^lny=i[lnt-ln(t+l)] => J £ = I (I - ^ Y t+i vt+i/ dy _ l / t (1 ]_' dt 2 V t + 1 \ t t+1, 1 / t 1 2 V t+1 |_t(t+l)J 2v/t(t+ 1)3/2 58. y = \/wh = W l + W~ 1/2 =* lny=|[lnt + ln(t+l)] 1 dy y dt 2 Vt T t+ W dy dt 1 / 1 2t + l 2 V t(t+l) t(t+l) 2t+l 2 (t 2 + 1) 3/2 59. y = y/6 + 3 (sin 0) = (0 + 3) 1 / 2 sin =^> In y = ± ln(0 + 3) + In (sin 9) => ± dv y d0 2(9 + 3) ' sin 1 I cos ( dy \/6 + 3 (sin 0) [ 5?? ^ + cot0 60. y = (tan 9) \/29 + 1 = (tan 0)(20 + 1) 1/2 =>■ In y = In (tan 0) + \ In (20 + 1) 1 dy _ sec 2 y d# — tan ( + (§)G£t) dy (tan 0) y/29 + 1 (^| + ^) = (sec 2 0) ^/20T 1 + tang 1 dy 1 1 . 1 y dt t ' t+1 ' t+2 61. y = t(t + l)(t + 2) => In y = In t + In (t + 1) + In (t + 2) t(t + i)(t + 2) r < t+ i)(t+2)+Kt+2)+t(t+ 1) |=t(t+l)(t+2)(i t ' t+1 ' t+2/ t(t+l)(t + 2) 3t 2 + 6t + 2 62 - y = t(t+ixt + 2) =* In y = in 1 - In t - In (t + 1) - ln(t + 2) =» -^ dy _ 1 r_ 1 _ 1 _ 1 I _ y dt t t+1 t + 2 i r_ i i i_] _ -l (t+i)(t+2)+t(t+2) + t(t+i) dt t(t+l)(t + 2) L t t+1 t + 2 J t(t+l)(t + 2) [ t(t+l)(t + 2) 3t 2 + 6t + 2 (t3 + 3t 2 + 2t) 2 63. y In y = In (6 + 5) - In 9 - In (cos 9) 1 dy _ _J_ _ 1 i sin£ y d8 ~ 9+5 ' cost S = (rK.)(jT3-i + '™») Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 436 Chapter 7 Transcendental Functions 64. y =^ In y = In 6 + In (sin 9) - \ In (sec 0) cot - i tan (9) /secfl ' "" ' 2 dy _ flsing /l d9 v^9 ^ 9 1 dy _ 1 I cos y A9 ~ I 6 ~*~ sin 9 fl (sec 9)(tan 9) 2 sec ( 65. y = f^ => lny = lnx+iln(x 2 +l)-fln(x+l) y' = ¥M\U (x+l) 2 / 3 x ~ x 2 +l 3(x+l) i_ _ 1 i x 2_ y x ~ x 2 + 1 3(x+ 1) (x+l) 11 => lny = i[101n(x+ l)-51n(2x+ 1)] => 1-5 ""• y \/ (2x+l)= "^ "* } i 1 '""'^ ' ^ ■ ,iu ^" i ^J ->' y x + i 2x+l / _ /(x+l) 10 / 5 f_5 2_) V x + 1 2x + 1 1 (2x+l) 5 Vx+1 2x + l 67 - y= V^+i => lny = j[ lnx + ln ( x -2)- ln ( x2 + 1 )] => i _ 1 3 / x(x - 2) /l i 1_ _ 2x ' y — 3 V x 2 + 1 Vx + x-2 x 2 + 1 , t — I (l 4. _J_ _ 2x \ y 3U T i-2 x 2 + W 68. y 3 / x(x+l)(x-2) ;x 2 + 1 ) (2x + 3) =4> In y = \ [In x + In (x + 1) + In (x - 2) - In (x 2 + 1) - In (2x + 3)] .,/ _ 1 3 / x(x+l)(x-2) /l , 1 . L_ 2x 3 V (x 2 + l)(2x + 3) Vx r x + l ^ x-2 x 2 + 1 2x + 3; 69. (a) f(x) = In (cos x) =^ f (x) sin X COS X tan x = =4> x = 0; f (x) > for - \ < x < and f (x) < for < x < | =>• there is a relative maximum at x = with f(0) = In (cos 0) = In 1 = 0; f (— |) = In (cos (— f )) = In I -4- ) = — j In 2 and f (|) = In (cos (|)) = In \ = — In 2. Therefore, the absolute minimum occurs at x = | with f (f) = — In 2 and the absolute maximum occurs at x = with f(0) = 0. (b) f(x) = cos (In x) =>• f'(x) = ~ sin x (lnx) =0 =4> x = 1; f'(x) > for \ < x < 1 and f'(x) < for 1 < x < 2 => there is a relative maximum at x = 1 with f(l) = cos (In 1) = cos 0= 1; f (I) = cos (in (|)) = cos (— In 2) = cos (In 2) and f(2) = cos (In 2). Therefore, the absolute minimum occurs at x = | and x = 2 with f (i) = f(2) = cos (In 2), and the absolute maximum occurs at x = 1 with f(l) = 1. 70. (a) f(x) = x — In x => f'(x) = 1 — - ; if x > 1, then f'(x) > which means that f(x) is increasing (b) f(l) = 1 - In 1 = 1 =>■ f(x) = x - In x > 0, if x > 1 by part (a) =^ x > In x if x > 1 71. J (In 2x - In x) dx = J (- In x + In 2 + In x) dx = (In 2) f dx = (In 2)(5 - 1) = In 2 4 = In 16 72. A = f ° - tan x dx + f* \an x dx = f ° ^^ dx - f* " ^^ dx = fin Icos xll ° w , , - fin |cos xll „ /3 J-jr/4 Jo J-tt/4 cosx Jo cosx LI IJ -jr/4 II IJ = (in 1 - In 4-) - (In | - In l) = In \fl + In 2 = | In 2 73. V = nf^-^y dy = 4^ £ ^ dy = 4^ [In |y + l|]g = 4tt(1ii 4 - In 1) = 4^ In 4 74. V = 7T P "cot x dx = 7T P ~ ^ dx = tt fin (sin x)l ^ = tt (in 1 - In ±) = tt In 2 J tt/6 J tt/6 sin X 1 V 'J 7T/G V 2 / 75. V = 2nf* x (i) dx = 2tt | J dx = 2tt [In |x|] j /2 = 2tt (in 2 - In ±) = 2tt(2 In 2) = tt In 2 4 = tt In 16 Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.2 Natural Logarithms 437 76. V = 7T jY-^i— ) " dx = 21 tt £ dx = 27tt [In (x 3 +9)f = 277r(ln 36 - In 9) = 277r(ln 4 + In 9 - In 9) = 27tt In 4 = 54tt In 2 77. ( i iM< = f-tax~l + (yT = l+(f-I)- = l+(%4) =(^) ^L=/ 4 7l + (y') 2 dx J^^p dx = J 4 8 (| + i) dx = [f + In |x|] = (8 + In 8) - (2 + In 4) = 6 + In 2 (b) x=(^-21n(|) =* | = I-| => l + [£ (*)' i ( s — f i =i +(^) 2 =(^y 8 + 2 In 3 = 8 + In 9 12 + 21ny| =(9 + 2 In 12) -(1 + 2 In 4) 78. L = Jj v/TT^i dx => -| = i =4> y = In |x| + C = In x + C since x>0 => = lnl + C =>• C = =^ y = lnx 79. (a) M y = /;x(i)dx=l,M x = /;(i)(i)dx=ij; 2 idx=[-i] 2 =I,M=j; 2 I ^dx= [lnlxllf =ln2 X — M _ In 2 ~ i "** dHU J — M — In 2 ~ ^^ (b) 80. (a) M y = f\ (-J-) dx = J/V/2 dx = f [ X 3/ 2 ] J 6 = 42; M, = /"(^ (-J.) dx = 1 /"i dx = 1 [In |x|]}° =ln4,M= /"^ dx = [2xW] \" = 6 => x = g = 7 and y = $ = ^ w ^ = r * fe) (^) dx = 4 /> = «>. m * = x 16 {&) (^) (^) dx = 2 /r x - 3/2 dx = -4[x-V 2 ]; c = 3,M = /;(-i ; )(^)dx = 4/;idx=[41n|x|]; 6 =41nl6=>x My _ J5_ M In 16 and M 4 In 16 1. ffl = 1 + I at (1, 3) => y = x + In |x| + C; y = 3 at x = 1 => C = 2 => y = x + In |x| + 2 dx- 1 dy tan x + C and 1 = tan + C =4> jj| = tan x + 1 =>■ y = J (tan x + 1) dx = In |sec x| + x + Ci and = In |sec 0| + + Ci => Ci = => y = In |sec x| + x 83. (a) L(x) = f(0) + f (0) - x, and f(x) = In (1 + x) =>• f ' (x)| x=0 = -^ | x=o = 1 =>■ L(x) = In 1 + 1 • x => L(x) = x (b) Let f(x) = ln(x + 1). Since f"(x) = — l 2 < on [0, 0.1], the graph of f is concave down on this interval and the (x+lj largest error in the linear approximation will occur when x = 0.1. This error is 0.1 — ln(l.l) « 0.00469 to five decimal places. Cfigl (c) 1 Pea Education, Inc., publishing as Pearson Addison-Wesle 438 Chapter 7 Transcendental Functions (c) The approximation y = x for In (1 + x) is best for smaller positive values of x; in particular for < x < 0.1 in the graph. As x increases, so does the error x — In (1 + x). From the graph an upper bound for the error is 0.5 - ln(l + 0.5) « 0.095; i.e., |E(x)| < 0.095 for < x < 0.5. Note from the graph that 0.1 - In (1 + 0.1) w 0.00469 estimates the error in replacing In (1 + x) by x over < x < 0.1. This is consistent with the estimate given in part (b) above. 0.5 y = x/ 0.4 / y' 0.3 / , ■■' y = In (> 0.2 /' 0.1 O.'l Q'l 013 0!4 0:5 x 84. For all positive values of x, 4- [ In | ] = i • — 53 = — j and ^ [ In a — In x ] = 0— 1 = — £ . Since In j| and In a — In x have x the same derivative, then In - = In a — In x + C for some constant C. Since this equation holds for all positve values of x, it must be true forx= 1 =4> In - =lnl — lnx + C = — lnx + C=>ln- = —In x + C. By part 3 we know that In - = -In x => C = =^ In - = In a - In x. 85. y = In kx =>- y = In x + In k; thus the graph of y = In kx is the graph of y = In x shifted vertically by In k, k > 0. 86. To turn the arches upside down we would use the formula y = — In jsin x| In 10 15 20 y ■ In I sin x| 87. (a) (b) y' Since Isin xl and Icos xl are less than a+sm x or equal to 1, we have for a > 1 ^r < y' < -Xr for all x. a— 1 — J — a— 1 Thus, lim y' = for all x ^> the graph of y looks a— >+oo more and more horizontal as a — > + oo. Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle (a) The graph of y = ^/x — In x appears to be concave upward for all x > 0. (b) y = Ay/x — In x => y' _j i 1 4x3/2 Section 7.3 The Exponential Function 439 + i = i(-^ + i) rs ar x => ,/x = 4 => x = 16. Thus, y" > if < x < 16 and y" < if x > 16 so a point of inflection exists at x = 16. The graph of y = y/x — In x closely resembles a straight line for x > 10 and it is impossible to discuss the point of inflection visually from the graph. 7.3 THE EXPONENTIAL FUNCTION 1. (a) e ln72 = 7.2 2. (a) e ln ( x2 +y 2 )=x 2 + y 2 (b) e- lnx2 -- (b) e- lnCO (c) e lnx ~ lny = e ln ' x / y ' = - 3. (a) 21n v /e = 21ne 1 / 2 = (2)(i) lne= 1 (c) lne(- x2 -y 2 > = (-x 2 - y 2 ) In e = -x 2 - y 2 4. (a) In (e secfl ) = (sec 6»)(ln e) = sec 6 (c) In (e 21nx ) = In (e lnx2 ) = In x 2 = 2 In x l j_ e ln0.3 03 (c) e ln7rx— In 2 pln(7rx/2) yrx (b) In (In e e ) = ln(e In e) = In e = 1 (b) In e( eX > = (e x ) (In e) = e x 5. In y = 2t + 4 ^ e lny = e 2t + 4 ^> y = e a 2t+4 6. In y = -t + 5 => e lny = e-'+ 5 => y = e~ t+5 7. ln(y - 40) = 5t ^ e ln ( y - 40) = e 5t => y - 40 = e 5t => y = e 5t + 40 8. ln(l - 2y) = t => e 1 "' 1 - 2 ^ = e l => 1 - 2y = e< => -2y = e< - 1 =>• y = - (^f 1 ) y-lx 9. In (y - 1) - In 2 = x + In x =>• In (y - 1) - In 2 - In x = x =^ In v , x , ^ =>■ y - 1 = 2xe x => y = 2xe x + 1 10. ln(y 2 - 1) -ln(y+ l) = ln(sinx) =^> In (y^j) = In (sin x) =>■ ln(y - 1) = In (sin x) =4- e 1 "^ 1 ) =^> y — 1 = sin x =>- y = sin x + 1 y-i a In (sinx) 11. (a) e 2k = 4 =>• In e 2k = In 4 => 2k In e = In 2 2 =^ 2k = 2 In 2 => k = In 2 (b) 100e 10k = 200 =>■ e 10k = 2 => In e 10k = In 2 =>• 10k In e = In 2 => 10k = In 2 =* k = ^ (c) e k / 1000 = a => In e k / 100() = In a ^ ^ In e = In a =>■ j^ = In a =^ k = 1000 In a 12. (a) e 5k = \ => In e 5k = In 4- 1 => 5k In e = - In 4 => 5k = - In 4 => k = - ^ (b) 80e k = 1 => e k = 80" 1 => In e k = In 80" 1 => k In e = - In 80 => k = - In 80 (c) e (in0.8)k = 08 ^ ( e lll0 - 8 ) k = 0.8 => (0.8) k = 0.8 => k=l Cf|t (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 440 Chapter 7 Transcendental Functions 13. (a) e-°- 3t = 27 => In e -03 ' = In 3 3 => (-0.3t) In e = 3 In 3 => -0.3t = 3 In 3 => t=-101n3 (b) e kt = 1 => In e kt = In 2- 1 = kt In e = - In 2 => t = - ^ (c) e (in0.2)t = 4 ^ (e 1 " 02 )' = 0.4 =S> 0.2 [ = 0.4 =4> In 0.2' = In 0.4 =>• t In 0.2 = In 0.4 => t = |^| 14. (a) e- u - uu =1000 =>• In e- uult = In 1000 => (-O.Olt) In e = In 1000 => -O.Olt = In 1000 =*> t= -100 In 1000 (b) e kt = i => In e kt = In 10" 1 = kt In e = - In 10 =» kt = - In 10 => t - ln 10 10 a (ln2)t _ 1 => e ln2\ 2- y => 2 t = 2~ I =* t= -1 (c) e^ 15. e^ /t = x 2 => lnev /t = lnx 2 => \/t=21nx => t = 4(ln x) 2 16. e x2 e 2x+1 =e l =>• e x2+2x+1 = e l ^ In e x 2 +2x+i = Jn e t ^ t = x 2 + 2x + 1 17. y = e~ 5x => y' = e~ 5x £ (-5x) => y' = -5e" 18. y = e 2x / 3 => y' = e 2x / 3 £ (f ) y' = | e 2x / 3 19. y = e 5 - 7x =>/ = e 5 ~ 7x £ (5 - 7x) => y' = -7e 5 - ?x 20. y = e( 4 v /x "+ x2 ) =>. y = e (V*"+* 2 ) £ (4^ + x 2 ) => y' = (4- + 2x) e ( 4 v /x "+ x2 ) 21. y = xe x — e x =4> y' = (e x + xe x ) — e x = xe x 22. y = (1 + 2x)e~ 2x ^ y' = 2e~ 2x + (1 + 2x)e~ 2x £ (-2x) => y' = 2e~ 2x - 2(1 + 2x)e~ 2x = -4xe~ 2x 23. y = (x 2 - 2x + 2) e x => y' = (2x - 2)e x + (x 2 - 2x + 2) e x = x 2 e x 24. y = (9x 2 - 6x + 2) e 3x => y' = (18x - 6)e 3x + (9x 2 - 6x + 2) e 3x £ (3x) ^ y' = (18x - 6)e 3x + 3 (9x 2 - 6x + 2) e 3 = 27x 2 e 3x 25. y = e fl (sin 9 + cos 9) => y' = e"(sin 9 + cos 6) + e"(cos 9 - sin 6) = 2e e cos 9 26. y = ln (39e-°) = ln 3 + ln 9 + ln e~ 8 = ln 3 + ln 6 - 9 => % = \ - 1 27. y = cos (e-* 2 ) =* % = - sin (e^) A ( e -* 2 ) = (_ sin (V« 2 )) (e^ 2 ) ± (-6 2 ) = 29^ sin (e^) 28. y = 9 3 e- w cos 56* => % = (3<9 2 ) (e" 26 cos 5(9) + (6 3 cos 56) t' 20 £ (-29) - 5(sin 59) (6 3 e- 20 ) = 9 2 e- 2e (3 cos 5(9 - 26 cos 59 - 56 sin 50) 29. y = ln (3te~ l ) = ln 3 + ln t + ln e~ l = ln 3 + ln t - t dy _ 1 1 30. y = ln (2e _t sin t) = ln 2 + ln e _t + ln sin t = ln 2 - t + ln sin t cos t — sin t dy dt \ sin t / dt (sin t) = -1 31. y = ln l+e" In e" - ln 1 + e? = 9 - ln l+e tf =>■ e\ -k ^y , l+e« l + e» =1 l+e» 1 l+e« Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.3 The Exponential Function 441 32. y = ln^ =lnV0-ln(l + Vfl) dy J9 (i + v / »)-\/e_ i i 20 (l + v^) ~ 29(l + v^) ~ 29(1+91/2) 33. y = e ( cost + lnt ) = e cost e lnt = te cost ^ % = e cost + te cost | (cos t) = (1 - t sin t) e cost 34. y = e sint (In t 2 + 1) => § = e sint (cos t) (In t 2 + 1) + \ e sint = e sint [(In t 2 + 1) (cos t) + f] 35. f * sin e l dt =>■ y' = (sin e lnx ) • A (In x) = sp 36. y = /j^ In t dt => y' = (In e 2x ) • A (e 2x ) - (in e 4 v^) - A ( e VZ) = (2x ) (2e 2x ) - (4^) (e 4 v^) - A (4^) = 4xe 2x - 4 v /xe 4l A (-2=) = 4xe 2x - Se 4 ^ 37. In y = e y sin x =^- I - ) y' = (y'e y ) (sin x) + e y cos x =4> y' I i — e y sin x) = e y cos x l ( 1 — ye y sin x \ 1 \ y / e y cos x => y / ye y cos x 1 — ye y sin x 38. lnxy = e x+y => In x + In y = e x+y => I + (i) y' = (1 + y') e x4 => y ' f i-y^ +i \ _ xe^+'-i _^ „/ _ y(xe*+>'-i) /( y -e x+y ) a x+y _ 1 x(l -ye"+y) 39. e 2x = sin(x + 3y) => 2e 2x = (1 + 3y') cos(x + 3y) => 1 + 3y' = ^f^ => 3y' = ^_ / _ 2e 2x - cos (x + 3y) ^ " 3cos(x + 3y) 40. tan y = e x + In x =* (sec 2 y) y' = e x + 1 =>■ y' = (xeX + ^ cos2 y 41. J(e 3x + 5e~ x ) dx = ^ - 5e~ x + C 42. J(2e x - 3e~ 2x ) dx = 2e x + § e~ 2x + C 43. f e x dx = [e x ]J^ = e ln3 - e ln2 = 3-2=1 44. J_" n2 e~ x dx = [-e- x ]°_ ln2 = -e° + e lnI 45. /8e( x+1 'dx = 8e( x+1 ' +C 46. J^e' 2 "- 1 ) dx = e^ 1 ' + C 47. Jj^V 2 dx = [ 2eX/2 ] ta4 = 2 [ e( ' n9)/2 " e(ln4>/2 ] = 2 ( e ' n3 - e ' n2 ) = 2(3 - 2) = 2 48. /* 16 e x / 4 dx = [4e x / 4 ] J" 6 = 4 ( e ( lnl6 >/ 4 - e°) = 4 (e ln2 - l) = 4(2 - 1) = 4 1+2=1 Let u = r 1 / 2 ^ du = i r" 1 / 2 dr => 2 du = r" 1 / 2 dr; J^ dr = / e rV2 • r- 1 / 2 dr = 2 J*e u du = 2e u + C = 2e fl/2 + C = 26^ + C 50. Let u = -r 1 / 2 => du = - \ r~ 1/2 dr =4> -2 du = r" 1 / 2 dr; f^£ dr = JV rl/2 • r- 1 /' 2 dr = -2 Je 11 du = -2 e - r ' /2 + C = -2e"^ + C Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 442 Chapter 7 Transcendental Functions 51. Letu=-t 2 => du = -2tdt => -du = 2t dt; J*2te- t2 dt = -Je u du = -e u + C = -e~ t2 + C 52. Let u = t 4 =>• du = 4t 3 dt => \ du = t 3 dt; J" 1 3 e t4 dt = \ Je u du = \ e t4 + C 53. Let u = - => du = - \ dx =4> -du = 4 dx; X X 2 X 2 ' J^ dx = J-e u du = -e u + C = -e'/ x + C 54. Let u = -x 2 =4> du = 2x 3 dx => | du = x -3 dx; J"^£- dx = JV X 2 • x- 3 dx = 1 Je u du = \ e u + C = \ e~ x 2 + C = \ e" 1 /" 2 + C 55. Let u = tan => du = sec 2 d0; = =>• u = 0, 9 = f =4> u=l; /^(l + e tanfl ) sec 2 d0 = J^ 'sec 2 d0 + /Je" du = [tan 0] l' A + [e u ] J = [tan (£ = (l-0) + (e-l) = e tan(0)] +(e : - e°) 56. Let u = cot => du = - esc 2 d0; = | =>- u = 1, /^ 2 (1 + e c ° tfl ) esc 2 d0 = £'l esc 2 d0 - J" e u du = (0+l)-(l-e) = e | ^ u = 0; -cot0]:g-[e"];=[-cot(f)+cot(|)]-(e°-e 1 ) 57. Let u = sec 7rt =>- du = 7r sec nt tan 7rt dt sec 7rt tan 7rt dt; J e sec (rt) sec (7rt) tan (7rt) dt = i Je u du = f + C = s-^ + C 58. Let u = esc (7T + t) =» du = - esc (n + t) cot {n + t) dt; f e csc (7r+t) csc (7r + t) cot (7r + 1 ) dt = - Je u du = -e u + C = -e csc fr+Q + C 59. Let u = e v => du = e v dv => 2 du = 2e v dv; v = In | => u = |, v = In f =>• u = §; J»ln (7r/2) r^/^ /O i„(,/6, 2e v cose v dv = 2j V6 cosudu=[2sinu]^ = 2[sin(|)-sin(|)]=2(l-I) = l 60. Let u = e x => du = 2xe x dx; x = =4> u = 1, x = V In 7r =>• u = e ln7r = 7r; J 2xe x cos I e x ) dx = J cos u du = [sin u] * = sin(7r) — sin(l) = — sin(l) w —0.84147 61. Let u = 1 + e r =>• du = e r dr; JlT? dr = Ji du = In |u| + C = In (1 + e r ) + C 62. f T -h r dx= f^ydx; J 1 + e 1 J e x + 1 ' let u = e~ x + 1 =>■ du = — e~ x dx => — du = e _x dx; J^TT dx = - / i du = - In |u| + C = - In (e- x + 1) + C 63. Jy e' sin (e ( - 2) =4> y = JV sin (e l - 2) dt; let u = e l - 2 =>• du = e' dt =4> y = J sin u du = - cos u + C = - cos (e l - 2) + C; y(ln 2) = Copyright (c| 1 Pen Etation, Inc., publishing as Pearson Addison-Wesle Section 7.3 The Exponential Function 443 => - cos (e ln2 - 2) + C = => - cos (2 - 2) + C = => C = cos = 1; thus, y = 1 - cos (e l - 2) 64. dy e ' sec 2 (7re ') => y = I e ' sec 2 (7re ') dt; let u = 7re ' =>• du = -7re 'dt =4> - ^ du = e ' dt =>• y = — - / sec 2 u du = - - tan u + C TT J TT J TT = - I tan (tkT 1 ) + C; y(ln 4) = § =* - i tan (™-' n4 ) + C = 2 => - ± tan (tt • i) + C = £ => -±(1) + C= ^ => C= ^;thus, y = 2 - I tan^e"') 65. S = 2e~ x ^ f = -2e~ x + C; x = and ^ = => 0= -2e° + C => C = 2; thus & = -2e- x + 2 ax 2 dx ' dx ' dx => y = 2e~ x + 2x + Ci; x = and y = 1 =^> 1 = 2e° + Ci => Ci = -1 => y = 2e~ x + 2x - 1 = 2 (e~ x + x) - 1 66. d 2 v dt- |=t-ie-+^-l ::- V I - c- 1 -> ^ = t ~ 5 e 2t + C; t = 1 and f = => = 1 - | e 2 + C => C = | e 2 - 1; thus a 2t _^ dy dt e 2 it 2 -ie 2 <+(ie 2 l) t + Ci;t = 1 andy i _ I ».2 i 1 -2 2 4 C " T " 2 Ci = -!- ip 2 v-1 2 4 y=it 2 -Ie 2t - '1 a 2 e 2 -l)t-(i + Ie 2 ) 67. f(x) = e x - 2x =4> f (x) = e x - 2; f (x) = =>■ e x = 2 => x = In 2; f(0) = 1, the absolute maximum; f(ln 2) = 2 — 2 In 2 w 0.613706, the absolute minimum; f(l) = e — 2 « 0.71828, a relative or local maximum since f "(x) = e x is always positive. 68. The function f(x) = 2e sln ' x /^ has a maximum whenever sin | = 1 and a minimum whenever sin | = — 1. Therefore the maximums occur at x = 7r + 2k(27r) and the minimums occur at x = 3tt + 2k(27r), where k is any integer. The maximum is 2e ss 5.43656 and the minimum is - w 0.73576. 69. f(x) = x 2 In i =4> f'(x) = 2x In i + x 2 (\\ (-x~ 2 ) = 2x In i - x = -x(2 In x + 1); f'(x) = => x = or In x = — i . Since x = is not in the domain of f, x = e~ 1//2 = -4= . Also, f '(x) > for < x < -4- and z v e v e f '(x) < for x > -j- . Therefore, f I -K- ) = \ In \fk = \ In e 1 / 2 = ^ In e = i is the absolute maximum value of f assumed at x 70. f(x) = (x - 3) 2 e =* f (x) = 2(x - 3) e x + (x - 3) 2 e x = (x - 3) e x (2 + x - 3) = (x - l)(x - 3) e x ; thus f (x) > for x < 1 or x > 3, and f'(x) < for 1 < x < 3 => f(l) = 4e w 10.87 is a local maximum and f(3) = is a local minimum. Since f(x) > for all x, f(3) = is also an absolute minimum. r( x ) = (x-l)(x-3)< 71. X"V x -e x )dx=[^ -e x ]" 3 =(^-e-) - (f - e°) = (| - 3) - (I - 1) = §-2 = 2 72. J o 2 '° 2 (e x / 2 - e- x / 2 ) dx = [2e x / 2 + 2e- x / 2 ] 2 In 2 2e ln2 + 2e- ln2 j - (2e° + 2e u ) = (4+ 1) - (2 + 2) = 5 - 4 = 1 73 - L = Xv T +? dx I = ?|2 => y = e x / 2 + C; y(0) = => = e° + C => C = -l => y = e x / 2 - 1 Cf|t (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 444 Chapter 7 Transcendental Functions 74. S = 2tt f Jo p In = 2tt Jo ' e' - + e~ ln2 /- e y + e ->\ 2 1 + \ (e 2 y - 2 + e- 2 y) dy dy= | J o " ,2 (e 2 5' + 2 + e- 2 >')dy ■n [I P.2y 2 L2 e'> + 2y _ 1 e -2yl ln2 2 C Jo |(i. 4 + 21n2-I-i) = |[(Ie 2 '« 2 + 21n2-Ie- 2 >« 2 )-(i+0-i)] |(2-I+21n2)=7r(if+ln2) 75. (a) -f (x ln x - x + C) = x • \ + ln x - 1 + = ln x (b) average value = ^y I ln x dx = ^rj [x ln x — x] *j = ~j [(e ln e — e) — (1 ln 1 — 1)] = J- r (e-e+l)=^- r e— 1 v / e — 1 76. average value = ^y I - dx = fin Ixll t = ln 2 — ln 1 = ln 2 77. (a) f(x) = e => f (x) = e x ; L(x) = f(0) + f'(0)(x - 0) =>• L(x) = 1 (b) f(0) = 1 and L(0) = 1 =>• error = 0; f(0.2) = e (u (c) Since y" = e x > 0, the tangent line approximation always lies below the curve y = e x Thus L(x) = x + 1 never overestimates e x . 1.22140 and L(0.2) = 1.2 y error w 0.02140 y = e I I 78. (a) e x e~ x = e( x - x ) = e° = 1 =$■ e" x = h for all x • 2*i = p x i(J- x l p— x 2 — f> x l~ x 2 e A1 e (b) y = (e Xl )" 2 =>• ln y = x 2 ln e Xl = x 2 Xi = XiX 2 => e In y — o x l x 2 X A X 2 _ „ x l x 2 (e Xl ) 79. f(x) = ln(x) - 1 => f (x) X n +1 ln (x„) - 1 x n+ i = x n [2 — In (x n )] . Then Xi = 2 =4> x 2 = 2.61370564, x 3 = 2.71624393 and x 5 = 2.71828183, where we have used Newton's method. 80. e lnx = x and ln (e x ) = x for all x > 81. Note that y = ln x and e y = x are the same curve; / ln x dx = area under the curve between 1 and a; pin a I e y dy = area to the left of the curve between and ln a. The sum of these areas is equal to the area of the rectangle =4> I ln x dx + I e y dy = a ln a. 82. (a) y = e x =>■ y" = e x > for all x =4> the graph of y = e x is always concave upward />lnb (b) area of the trapezoid ABCD < J e x dx < area of the trapezoid AEFD => ± (AB + CD)(ln b - ln a) < J e x dx < ( e "'" + e '" b j (ln b - ln a). Now \ (AB + CD) is the height of the midpoint M = e ( llla+lnb '/ 2 since the curve containing the points B and C is linear =>• e (lna+lnb ^ 2 (ln b — ln a) < £" b e x dx < ( e "'" + e "' Ll ) (ln b - ln a) f hlb tab (c) I e x dx = [e x ] ° a = e lnb — e lna = b — a, so part (b) implies that e (ina+inb)/2 (i n b - In a)< b - a < ( e '" + e ' nb ) (ln b - ln a) => e ( lna+lnb '/ 2 < ln !g:f na < S±* Vab < „ln a/2 „lnb/2 ^ b-a , a + b C ' C ^ ln b - ln a ^ 2 /pin a ,/plnb ^ b-a ~ a + b C V C ^ ln b - ln a ^ 2 b — a ^- a + b lnb-lna ^ 2 Copyright (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.4 a x and log a x 445 7.4 a" and log a x 1. (a) 5 l0&7 = 7 (b) 8 l0& (d) log 4 16 = log 4 4 2 = 2 log 4 4 = 2-1=2 (f) log 4 (1) =log 4 4- 1 = -llog 4 4 = -l-l = -l * = Ji (c) 1.3 log '-' 75 = 75 (e) log 3 y/3 = log 3 3 1 ' 2 = | log 3 3 = 4-1 0.5 2. (a) 2 l0fc3 = 3 (b) 10 loe '°( I/2 » = \ (d) log,, 121 = log,, 1 1 2 = 2 log,, 11=2-1=2 (e) log 121 11 = log, 2 , 121 1 / 2 = (I) log 121 121 = (I) - 1 = \ (f) log 3 (i) = log 3 3- 2 = -2 log 3 3 = -2 - 1 = -2 3. (a) Let z = log 4 x => 4 Z = x => 2 2z = x =4> (2 Z ) 2 = x =>■ 2 Z = v /x (b) Let z = log 3 x =4> 3 Z = x => (3 Z ) 2 = x 2 =^ 3 2z = x 2 => 9 Z = x 2 (c) log, (e('° 2 ) si " x ) = log 2 2 si,,x = sinx 4. (a) Let z = log 5 (3x 2 ) =>■ 5 Z = 3x 2 => 25 z = 9x 4 (b) log, (e x ) = x (c) log 4 (2 e * sinx ) = log 4 4< e " sinx )/ 2 = ^p ( C ) 7T k,g ' 7 = 7 c /„\ log2 x lnx _^_ ln_x lnx ln_3 ln_3 J ' W log 3 x ~~ In 2 ■ In 3 ~~ In 2 " lnx ~~ In 2 ( r \ logx a In a . In a In a . In x 2 2 In x ^ l°g x 2 a In x ' In x 2 In x In a In x f> ('a , l IPgi x lnx _^_ lnx lnx In 3 1 °- W log 3 x ~~ In 9 ■ In 3 ~~ 2 In 3 ' lnx ~~ 2 (U) l0 ^' = fax ^_ lnx lnx (I) ln2 Jn2_ W logyjx i n /Jo • to ,/2 (l)lnio' lnx ln 10 (h\ log2 x l2_x _;_ lnx _ lnx ln_8 3 ln 2 _ o W logg x — ln 2 • ln 8 — ln 2 " ln x — ln 2 — J (C) log a b _ lnb _^ lnji _ lnb _ lnb _ / ln b \ 2 logb a ln a ' ln b ln a ln a V ln a / 7. 3 log1 (7) + 2 log2 ( 5 > = 5 log5 W =>. 7 + 5 = x => x = 12 8. 8 logs (3) - e ln5 = x 2 - 7 log7 < 3x > => 3 - 5 = x 2 - 3x => = x 2 - 3x + 2 = (x - l)(x - 2) => x = 1 or x = 2 9. 3 log1 < x2) = 5e lnx - 3 • 10 loglu ^ => x 2 = 5x - 6 => x 2 - 5x + 6 = => (x - 2)(x - 3) = => x = 2orx = 3 10. ln e + 4- 21og " W = I i O g 10 100 => 1 + 4 log ^ x 2 > = i log 10 10 2 =^ 1 + x" => x 2 - 2x + 1 = => (x - l) 2 = => x = 1 11. y = 2" =>• y' = 2 s ln 2 12. y = 3" x 13. y = 5^ s =» | = 5^ (In 5) (1 s- 1 /' 2 ) = (|l|) 5^ 14. y = 2 s2 => f = 2 s2 (ln 2)2s = (ln 2 2 ) (s2 s2 ) = (ln 4)s2 s2 (2) =► 1 + A - f = y ' = 3- x (ln3)(-l) = -3" x in3 15. y = x^ =>• y' = 7TX ' = TTY^-') 16. y = t'- e =>• i=d-0r 17. y = (cos 0)^ dv V2 (cos 6»)( v5 "') (sin 6») Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 446 Chapter 7 Transcendental Functions 18. y = (In &=>% = 7r(ln 9)^ (I) = ^f^ 19. y = 7 secS In 7 20. y = 3 lme In 3 dy dy (7 secS In 7) (In 7)(sec (9 tan 9) = 7 secS (ln 7) 2 (sec 6 tan 0) (S"" 1 " In 3)(ln 3) sec 2 9 = 3 ,imS (ln 3) 2 sec 2 21. y = 2 sin31 => | = (2 si ° 3 ' ln2)(cos3t)(3) = (3cos3t)(2 sin3l )(ln2) 22. y = 5- cos2 ' => f = (5- cos2l ln5)(sin2t)(2) = (2sin2t)(5- cos2, )(ln5) 23. y = log 2 50 = |f =* S = (&) (i) (5) = lA* 24. y = log_, (1 + 01n3) ln(l+flln3) In 3 dy unMl+kl)^) 1 + In 3 25 v = — + — Z - J - y In 4 T In In x i In xf In x i ^ In x o In x ^ _J 3 4 ~~ In 4 "^ Z In 4 — J In 4 =? " ~~ x In 4 x In e In x x In x 2 6- y = m - S = ±5 - ^ = (21b) (* - ln *) => y' = (dr?) (1 - i) 27. y = log 2 r.log 4 r= (£§) (£f) 28. y = log 3 r.log 9 r=(^)(^) ln 2 r =>• dy (lr 2)(ln 4) dr ln 2 r =>■ dy (li 3)(ln 9) dr [(I^k4j]( 21nr )(7) x- 1 2xln5 21nr r(ln 2)(ln 4) 29- y = bfc((HS) M ) WHi) (In 3) In ( In 3 -2 In 3 ^Z = [ 1 1 (2 In r) CM = 21nr dr [ (In 3)(ln 9) J ^ ln l ' \t) r(ln 3)(ln < ln(|±i) =ln(x+ l)-ln(x- 1) dy _ 1 _ 1 dx — x+1 x-1 ~~ (x+l)(x-l) 30- y = log 5 V(3TT2) = 10 ^ (5TT I I tx \(ln5)/2 1„5 , , 7x ^(to5)/2 _ In [jZfry In 5 :¥) i?Mi) = | In 7x - \ In (3x + 2) ^ 31. y = sin(log 7 0) = sin (£|) Jv J _ (3x + 2)-3x dx 2-7x 2-(3x + 2) ' 2x(3x + 2) x(3x + 2) % = M&) - [cos (hi)} f_L_ \ In 77 J Vein 7 = sin 1 In MM 2 m V 3x + 2 J (log, 0) + jpj cos (log, 0) 32. y = log, (^f^) in#cos#\ _ In (sin g) + In (cos g) - In e" - In 2" _ In (sin g) + In (cos t I In 2 In 7 ss = r ■ w 7. " r s ivi ,i - v^i - rl = (rM (cot - tan 9 - 1 - In 2) dy (sin 0)(ln 7) (cos 0)(ln 7) In 7 In 7 V In 7 / v ' 33. y = log 5 e*=^ = ^ => y'- 34. y = log 2 f x 2 e 2 \ _ In x 2 + In t V2 % /xTiy " In 5 - In 2 - In %/x + 1 21nx + 2-ln2-lln(x+l) In 2 4(x+l)-x In 2 3x + 4 ^ y xln2 2(ln2)(x+l) 2x(x+l)(ln2) 2x(x+l)ln2 35. y = 3 l0fcl = S^'VC" 2 ) => g = [3 (tal V('» 2 )(ln 3)] (^ = 1 (log 2 3) 3 lusl 36. y = 3 log 8 (log, t) _ i 3 1n(log 2 t) _ 3 In (^) dy _ / J_ In 8 dt V In 8 I \ (In t)/(ln 2) V t In 2 I t(ln t)(ln 8) t(ln t)(ln 2) Copyright (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.4 a x and log a x 447 37. y=l0g 2 (8t'° 2 ) = 1 " 8 + W t '" 2 ) = 31n2 + (ln2)(lnt) = 3 + ln t _ dy _ 1 In 2 ln2 dt t tin ((e'"T° 38- Y = ^3^ = ^ = ^^T 1 = tsint =* f = sint + tcost 39. y = (x + l) x => ln y = ln (x + l) x = x ln (x + 1) =>• y - = ln (x + 1) + x • ^-^ => y' = (x + l) x [^ + ln (x + 1) 40. y = x< x+1 > => ln y = ln x< x+1 > = (x + 1) lnx =>■ y - = ln x + (x + 1) (±) = ln x + 1 + \ => y> = x (x+1 » (1 + i+lnx) 41. y= (0)'= (t 1 / 2 )'^ 2 => lny = lnt'/ 2 = (±) ln t => i$ = (§)(lnt)+'^ '^ - lnt ' : v2y Vt/ 2 '2 l = (^)'(1 t + i) 42. y = t^ = t< ll/2 ) => ln y = ln t^ = (t 1 / 2 ) (ln t) => 1 | = (1 I" 1 / 2 ) (i n t) + t 1 / 2 (I) = ^ dt f lnt+2 \ tN /t V 2\A / 43. y = (sin x) x =>• ln y = ln (sin x) x = x ln (sin x) =4> - = ln(sin x) + x (^S 5 ^) =4> y' = (sin x) x [ln (sin x) + x cot x] 44. y = x s,nx =>• ln y = ln x smx = (sin x)(ln x) =>• - = (cos x)(ln x) + (sin x) . / _ sinx f s in x + x(ln x)(cos x) ' X \ sin x + x (ln x)(cos x) 45. y = x lnx , x > => ln y = (ln x) 2 => y j = 2(ln x) Q) => y' = (x" lx ) ( ] -^f) 46. y = (ln x) lnx => ln y = (ln x) ln (ln x) =>?= ( ln(ln x x) + 1 )(lnx)'" x ;i)ln(lnx) + Onx)(^)i(lnx) ln(lnx) , 1 47- J*5 x dx=^+C 49 - fo**M=fo{k)'M 48. J(1.3) x QX — In (1.3) +( - M In \ n (l) 2(lnl-ln2) 2 ln 2 (*0 n0 a 5o - Jj- 9 m=L& M In In TIT (1-25) -24 _ 24 In 1 - ln 5 ~~ ln 5 51. Let u = x 2 => du = 2x dx =>■ | du = x dx; x=l =>• u=l,x= \J 2 =>■ u = 2; / ^ x2) d X =/;(i)2"du=H^]? = (^2)(2 2 -2 l In 2 52. Let u = x 1 / 2 ^> du = i x" 1 / 2 dx => 2 du dx . ;x=l =4> u=l,x = 4 =4> u = 2; 2 r $ dx = r 2xi/2 • *- i/2 dx = 2 r 2u du = i 2 ^] x = (A) ( 23 - 2 2 ) 4 ln 2 Cf|t (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 448 Chapter 7 Transcendental Functions 53. Let u = cos t =^ du = - sin t dt => - du = sin t dt; t = =>- u = 1, t = 5 => u = 0; 6 In 7 £ /2 7"« sin t dt = -X° 7" du = [- £] J = (-1) (7° - 7) 54. Let u = tan t => du = sec 2 1 dt; t = =>■ u = 0, t = t => u=l /; /4 (i) imi sec 2 tdt=j; ) a) u du= In 3 J V 3 J V 3 J — 3 In 3 55. Let u = x 2x => In u = 2x In x => ± g = 2 In x + (2x) Q) => jjs = 2u(ln x + 1) =>■ \ du = x 2x (l + In x) dx; x = 2 => u = 2 4 = 16, x = 4 => u = 4 8 = 65,536; J»4 r»65,536 ,_ „, x*(l + In x) dx = I J du = 1 [u] i6 - = | (65,536 - 16) 56. Let u = In x => du = * dx; x = 1 =>• u = 0, x = 2 =4> u = In 2; 1 \ ol»2 _ O0\ _ 2'° 2 -l 65,520 2 32,760 In 2 58. Jx( v/5 " 1 )dx= 4j+C 59. /; (^2 + l) x^ dx = [x(^)] * = 3 (^) 60. f>^ <*=[£] I = *fe^ = f=* = £ 61 . / ^ d* = /(£*>) (I) dx; [u = In x=> du= I dx ] - J(n^)(Ddx= In ^/udu=( i ^)(Iu 2 ) + C=^ + C 62. J] ^ dx = J] (|§) (i) dx; [u = In x => du = ± dx; x = 1 => u = 0, x = 4 => u = In 4] -r(^)(Ddx = r\^)udu=(^)[Iu 2 ]r=(^)[I(ln4) 2 ]=f^ = ^ = ln4 63- J^ 3 ^ dx = //(i^) (£f ) dx = /> dx = [I (In x) 2 ] J = I [(In 4) 2 - (In I) 2 ] = \ (In 4) 2 = \ (2 In 2) 2 = 2(ln 2) 2 64. j; e 21nl0 f s '" x) dx = p^fiP (i) dx = [(In x) 2 ]; = (In e) 2 - (In I) 2 = 1 65 ■ £^M^ ** = & T Mx + 2)](^) dx = (£) [filfe±ae] * = (£)[ J_\ (In 4) 2 _ (In 2)" vln 2/ 2 2 1 \ r4(ln2) 2 (ln2) 2 1 _ 3 In 2 /'10 1/10 login (lOx) ^„ _ 10 dx 10 \ [4(ln 10) 2 In 10 X / Jln(10x)](^)dx=( i ^)[M^)!] i to /10 ' 10 \ (In 100) 2 _ (In 1)' vln 10/ 20 2 vln 10/ 20 2 In 10 67 . r;2 j£f ^ dx = _ 2 _r; ln(x + 1)( _L T)dx _. , , r^ n,r_.^,[^_.^ = ln 10 68. /;^^>dx=^X 3 ln( X -l)(^ l) dx - Vln; 2 \ | (ln(x-l)) 2 2 \ I (In 2) 2 (In l) 2 In II 2 2 In 2 Copyright |c) 1 Pen Education, k, publishing as Pearson Addison-Wesle Section 7.4 a x and log a x 449 69 - I^t-, = I (%r) Q) dx = d" 10) / (£) (i) dx; [u = tax => du = 1 dx] -» (Inl0)/( I ^) (i) dx = (tal0) / ±du = (lnl0)ln|u|+C = (lnl0)ln|lnx|+C 70. f -7^2 = f -£ti = On 8) 2 f 2^ dx = (In 8) 2 S 1 ^ + C = - ^ + C J x(log N x) 2 J „ lo yi v 7 J x v 7 -1 lnx A Un 8 / 71. J^i dt= [In |t|]j nx = ln|lnx| -In 1 = In (lnx), x > 1 72. f* \ dt = [In |t|] i = In e x - In 1 = x In e = x 73. / L ' /X i dt= [ln|t|]I /x =ln |±| - In 1 = (In 1 - In |x|) - In 1 = -lnx, x >0 74. J- f X i dt = [J- In Itll x = lnx ^ |nl = l g X, X > In a J i t Lin a I U l In a In a fea ' 75. A = f_ 2 j^ dx = 2/ ^j dx; [u = 1+ x 2 => du = 2x dx; x = => u = 1, x = 2 => u = 5] -► A = 2 J' I du = 2 [In |u|] \ = 2(ln 5 - In 1) = 2 In 5 76. A = J^ 2('- x) dx = 2 J^ (i) x dx = 2 kD 2 n -2-) C_i) = J_ ln2M 2^ ln2 77. Let [H 3 + ] = x and solve the equations 7.37 = — log 1() x and 7.44 = — log| x. The solutions of these equations are 1(T 7 - 37 and 1CT 744 . Consequently, the bounds for [H 3 0+] are [1(T 744 , 1CT 7 - 37 ] . 78. pH = - log K) (4.8 x 1(T 8 ) = - (log 10 4.8) + 8 = 7.32 79. Let O = original sound level = 10 log 10 (I x 10 12 ) db from Equation (6) in the text. Solving O + 10 = 10 log 10 (kl x 10 12 ) fork => 10 log 10 (I x 10 12 ) + 10 = 10 log 10 (kl x 10 12 ) => log 10 (I x 10 12 ) + 1 = log, (kl x 10 12 ) => log 10 (I x 10 12 ) + 1 = log 10 k + log 1() (I x 10 12 ) => 1 = log,„ k => 1 = ££ => In k = In 10 =4> k = 10 80. Sound level with 101 = 10 log 10 (101 x 10 12 ) = 10 [log 10 10 + log 10 (I x 10 12 )] = 10+10 log 10 (I x 10 12 ) = original sound level + 10 => an increase of 10 db 81. (a) If x = [H 3 0+] and S - x = [OH"] , then x(S - x) = 10~ 14 => S = x + ^ => jjf = 1 - ^ d 2 S _ 2-10- 14 and d\- > => a minimum exists at x = 10 (b) pH=-log 10 (10- 7 )=7 (c\ [° H ~] - S^x W r H ,0+] - x (x+^H)-x 10" the ratio L Q l equals 1 at x = 10 -7 82. Yes, it's true for all positive values of a and b: log a b = p* 2 and log b a = pr =^> p^- = ^ = log., b Cf|t (c) 1 Pea Education, Inc., publishing as Pearson Addison-Wesle 450 Chapter 7 Transcendental Functions 83. From zooming in on the graph at the right, we estimate the third root to be x w —0.76666 X-- 0.76666 84. The functions f(x) = x ln2 and g(x) = 2 hlx appear to have identical graphs for x > 0. This is no accident, because x In2 = e lr l n2 \lnx (e ln2 ) 85. (a) f(x) = 2 X ^> f'(x) = 2 X In 2; L(x) = (2° In 2) x + 2° = x In 2 + 1 « 0.69x + 1 (b) / y = (ln2)i + l -3 7T—A li 1 i 1 1.4 // 1.2 // / j/ = (ln2)x + l y = 2* •7 /0.8 0.6 0.4 0.2 \ -0.5 o ■ ■ 0:3 ■ ■ -I x 86. (a) f(x) = log 3 x => f '(x) = ^ , and f(3) = £§ => L(x) = 3^ (x- 3) +£§ = 3^-^ + 1 w 0.30x + 0.09 (b) 87. (a) log 3 8 In 8 to 3 1.89279 (c) log 2() 17 = %% » 0.94575 (e) In x = (log 1() x)(ln 10) = 2.3 In 10 « 5.29595 (g) In x = (log 2 x)(ln 2) = -1.5 In 2 « -1.03972 88. (a) ^•log 1 „x=>^. 1 ^ ) = ^=log 2 x > 1.2 ^(x-Si-M^ 1.1 y/ y = log 3 x 1 0.9 0.8 //' 0.7 // / 2 2:5 3 3:5 4 * (b) log 7 0.5 = ^ (d) log„, 7 = £fc -0.35621 -2.80735 (f) In x = (log 2 x)(ln 2) = 1.4 In 2 w 0.97041 (h) In x = (log 10 x)(ln 10) = -0.7 In 10 « -1.61181 In a ]„„ „ _ ln_a lnjc _ ln_x _ 1 In b In a In b lu & b A (b) ^-10g a X: 1-2 dx V 2' k) = -xandA(i nx + c ) = 1. Since — x ■ - = — 1 for any x^0, these two curves will have perpendicular tangent lines. 90. e ln x = x for x > and ln(e x ) = x for all x Copffigl (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.5 Exponential Growth and Decay 451 91. Using Newton's Method: f(x) = ln(x) - 1 => f (x) = 1 =4> x n+1 = x n - ln( *f 1 =>■ x n+1 = x n [2 - ln(x n ; Then, Xi = 2, x 2 = 2.61370564, x 3 = 2.71624393, and x 5 = 2.71828183. Many other methods may be used. For example, graph y = In x — 1 and determine the zero of y. 92. (a) The point of tangency is (p, In p) and m tange nt = - since £ = j. The tangent line passes through (0, 0) =>• the equation of the tangent line is y = -x. The tangent line also passes through(p, In p) => In p = -p = 1 =>■ p = e, and the tangent line equation is y = ^x. (b) jp = — \ for x^0=^y = lnxis concave downward over its domain. Therefore, y = In x lies below the graph of y = -x for all x > 0, x ^ e, and In x < - for x > 0, x ^ e. j e ' / e / (c) Multiplying by e, e In x < x or In x e < x. (d) Exponentiating both sides of In x e < x, we have e' nxe < e x , or x e < e x for all positive x/e. (e) Let x = 7r to see that if < e n . Therefore, e 51 is bigger. 7.5 EXPONENTIAL GROWTH AND DECAY 1. (a) y = y e kl => 0.99y = y e loook => k = l -^ « -0.00001 (b) 0.9 = e ( - 000001)1 => (-0.00001)t = ln(0.9) => t = _'g^ w 10,536 years (c) y = y e( 20 ' 00 °) k w y e- (U = y (0.82) => 82% 2. (a) % = kp => p = p e kh where p = 1013; 90 = 1013e 2Ok =>■ k = '"(*>) -to (ioi3) a -0.121 (b) p = 1013e- 6 - 05 « 2.389 millibars (c) 900 = 1013e(-°- 12 ') h =>. -0.121h = In ($£) => h = ln ffl j ^ w 0.977 km 3. I = -0.6y => y = yoe" - 61 ; y = 100 => y = lOOe" - 61 =>• y = lOOe" 06 w 54.88 grams when t = 1 hr 4. A = A e kl =^> 800 = 1000e lok => k = "^ =>• A = 1000e (ln(,, - 8 >'' 10 >, where A represents the amount of sugar that remains after time t. Thus after another 14 hrs, A = 1000e (ln(0 - 8)/10)24 w 585.35 kg 5. L(x) = L e- kx =>• ^ = L e- 18k =4> In | = -18k => k=^« 0.0385 =4> L(x) = L e- 00385x ; when the intensity is one-tenth of the surface value, ^ = L e-° 0385x => In 10 = 0.0385x => x « 59.8 ft 6. V(t) = Voe"'/ 40 =4> 0.1V =V e- 1 / 40 when the voltage is 10% of its original value =>■ t= -40 In (0.1) w 92.1 sec 7. y = y e kl and y = 1 =>• y = e kl => at y = 2 and t = 0.5 we have 2 = e°- 5k => In 2 = 0.5k =>■ k = £f = In 4. Therefore, y = e( ,n4 > =>■ y = e 241n4 = 4 24 = 2.81474978 x 10 14 at the end of 24 hrs 8. y = y e kl and y(3) = 10,000 => 10,000 = y e 3k ; also y(5) = 40,000 = y e 5k . Therefore y e 5k = 4y e 3k =>. e 5k = 4e 3k => e 2k = 4 =4> k = In 2. Thus, y = yoe('° 2 »' =4> 10,000 = y e 31n2 = y e ,n8 =4> 10,000 = 8y ^ yo = io|00 = 1250 9. (a) 10,000e k « = 7500 => e k = 0.75 => k = In 0.75 and y = 10,000e< In,, - 75 \ Now 1000 = 10,000e( lnO - 75 >' In 0.1 = (In 0.75)t => t = jjj^L ~ g Q0 years (to the nearest hundredth of a year) 1 year) In 0.75 , n T^'it -v. t — _ In 0.75 (b) 1 = 10,000e (ln0 - 75 > => In 0.0001 = (In 0.75)t => t = lj \W^ ~ 32.02 years (to the nearest hundredth of a Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 452 Chapter 7 Transcendental Functions 10. (a) There are (60)(60)(24)(365) = 31,536,000 seconds in a year. Thus, assuming exponential growth, P = 257,313,431e kl and 257,313,432 = 257,313,431e( 14k / 31 ' 5,6 ' 00 °) => In ( l|fj|ff ) = M^fooo =4> k« 0.0087542 (b) P = 257,313,431e (00087542) ( 15 ' ss 293,420,847 (to the nearest integer). Answers will vary considerably with the number of decimal places retained. 11. 0.9Po = Poe k => k = In 0.9; when the well's output falls to one-fifth of its present value P = 0.2P => 0.2P = P e('° M > => 0.2 = e( lnM > => In (0.2) = (In 0.9)t ^ t = |§f « 15.28 yr 12- (a) d i 100 P r(x) 2000 f = - yig dx => lnp = - T i 5 x + C =► p = e(-°- 0l *+V = e c e -o.oix = Cie -o.«»ix. p(100) = 20.09 => 20.09 = Cie ( -°- 01)(100) => Cj = 20.09e rs 54.61 => p(x) = 54.61e-° 0,! ' (in dollars) (b) p(10) = 54.61e(-°- 01 »( 10 ) = $49.41, andp(90) = 54.61e ( - ,u,1 '( 90 ) = $22.20 (c) r(x) = xp(x) => r'(x) = p(x) + xp'(x); p'(x) = -.5461e- ,u,lx => r'(x) = (54.61 - .5461x)e-°- 01s . Thus, r'(x) = => 54.61 = .5461x => x = 100. Since r' > J6CK for any x < 100 and r' < for x > 100, then r(x) must be a maximum at x = 100. 50 100 150 200 13. (a) A e(°- 04 ) 5 = A e - 2 (b) 2A = A e (0 - 04 > => In 2 = (0.04)t => t = -£1 w 27.47 years In 2 0.04 17.33 years; 3A = A e (0 - 04)l ^ In 3 = (0.04)t 14. (a) The amount of money invested A after t years is A(t) = A e' (b) If A(t) = 3A , then 3A = A e' => In 3 = t or t « 1.099 years (c) At the beginning of a year the account balance is Aoe', while at the end of the year the balance is Aoe (t+1) . The amount earned is Aoe (1+1 ' — Aoe 1 = Aoe'(e - 1) « 1,7 times the beginning amount. 15. A(100) = 90,000 => 90,000 = lOOOe'C 00 ' => 90 = e l0,,r => In 90 = lOOr => r = ^ ~ 0.0450 or 4.50% 16. A(100) = 131,000 => 131,000 = 1000e 100r => In 131 = lOOr => r 17. y = yoe -0181 represents the decay equation; solving (0.9)yo = yoe -0 - 181 ^ w 0.04875 or 4.875% => t In (0.9) -0.18 0.585 days 18. A = A e kl and i A = A e 139k => \ = e lwk => k = "^ « -0.00499; then 0.05 A = Aoe- - 0049 " 139 In 0.05 -0.00499 600 days 19. y = y e" yoe- -(k)(3/k) Yoe" -3 _ yo ^ yo < In = (0-05)(yo) => after three mean lifetimes less than 5% remains e 3 ^ 20 20. (a) A = A e- k ' => \ = e" 2 - 645k ^ k = ^ « 0.262 (b) 1 w 3.816 years (c) (0.05)A = Aexpi In 2 2.645 t) => _i n20 =(-^)t => t 2.645 In 20 In 2 11.431 years 21. T - T s = (T - T s )e- k ', T = 90°C, T s = 20°C, T = 60°C => 60 - 20 = 70e-' 0k => | = e" 10 0.05596 Cf|t (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.6 Relative Rates of Growth 453 (a) 35 - 20 = 70e" ' 0559st => t w 27.5 min is the total time =>■ it will take 27.5 - 10 = 17.5 minutes longer to reach 35°C (b) T - T s = (T - TJe" 1 ", T = 90°C, T s = -15°C => 35 + 15 = 105e- W)559 <" => t « 13.26 min 22. T - 65° = (T - 65°)e- kl => 35° - 65° = (T - 65°) e" 10 " and 50° - 65° = (T - 65°)e- 20k . Solving -30° = (T - 65°)e- 10k and -15° = (T - 65°)e- 20k simultaneously => (T - 65°) e" 101 ' = 2(T - 65°)e- 20k => e 10k = 2 => k = ^ and -30° = ^^ => -30° [e 10 ^) ] = T - 65° => T = 65° - 30° (e 1 " 2 ) = 65° - 60° = 5° 23. T - T s = (T - T s ) e" 1 " => 39 - T s = (46 - T s ) e" 10k and 33 - T s = (46 - T s ) e" 20k => 20k _^ 39-T, 46-T, e" 10k and / p -i0k\2 33-T, _ / 39-T, l c ) =" 46-T ~~ \ 46-T, 33-T, _ 20k 46-T s ~~ c = 1521 - 78T S + T 2 => -T s = 3 => T s = -3°C (33 - T s )(46 - T s ) = (39 - T s ) 2 =>■ 1518 - 79T S + T 2 24. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room temperature the silver will be 120 min from now, and to the time the silver will be 10°C above room temperature. We then have the following time-temperature table: time in min. 20 (Now) 35 140 to temperature T s + 70° T s + 60° T s + x T s + y T s + 10° T - T s = (T - T s ) e- kl => (60 + T s ) - T s = [(70 + T s ) - T„] e" 20k => 60 = 70e- 20k => k = (- ^) In (f ) w 0.00771 (a) T - T s = (T - T s )e- ,U)077U =^> (T, +x) -T, = [(70 + T.) - T,] e - ((u,0771 >( 35 ) =^ x = 70e" (b) T - T s = (T - TJe-"- 007711 => (T, + y) - T s = [(70 + T.) - Tj e -( - ,,0771 )( 14 ") ^ y = 70e (c) T - T s = (T - T s )e- ,U)077U => (T, + 10) - T, = [(70 + TJ - T,] e -<°- 00771 ) l ° =>• 10 = 70e- ,u,077Uil o.269 85 _ 53.440c -1.0794 r^, jo 7Q°r 1 In -0.00771t => t 7 y ^.w«,, iH , —r hj y 0.00771 silver will be 10°C above room temperature In 252.39 => 252.39 - 20 « 232 minutes from now the 25. From Example 5, the half-life of carbon-14 is 5700 yr c = coe (0.445)c = c e- i c = c e- k ( 57 ™) In (0.445) -0.0001216 k= ±± re 0.0001216 6659 years 26. From Exercise 25, k re 0.0001216 for carbon-14. (a) c = coe-"- 00012161 => (0.17)c = coe-"- 00012161 => t re 14,571.44 years (b) (0.18)c = coe- - 00012161 => t re 14,101.41 years => 12,101 BC (c) (0.16)c =c e- 000012 "" =>• t re 15,069.98 years =>■ 13,070 BC 12,571 BC 27. From Exercise 25, k re 0.0001216 for carbon-14. Thus, c = Coe In (0.995) -0.0001216 41 years old (0.995)c = c e- 7.6 RELATIVE RATES OF GROWTH 1. (a) slower, lim ^^ — \i m J_ = q x ^ oo e x ^ oo e (b) slower, lim * 3 + f 2 * = lim 3 " 2 + 2 g x cos x = lim 6x + 2 x cos2x = lim 6 ~ 4 f n2x = by the x ^ oo e x ^ oo e x ^ oo e x ^ oo e J Sandwich Theorem because 4 < 6 ~ 4 ' in2x < ^ for all reals and lim 4=0= lim *Q (c) slower, lim -v = lim ^ = lim X ^ OO e X — > OO e x ^ oo (d) faster, lim § = lim X — > CO e X — > oo e x f4\ x § s- 1/2 1 lim , - x — > oo V xe oo since - > 1 (e) slower, lim ^f- = lim (^ v y X — » OO e X — > oo v 2e ) x = o since ^ < 1 Cf|t (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 454 Chapter 7 Transcendental Functions (f) slower, lim X — > oo lim OO e x . (h 2. (a (b (c (d (c (f) (g (h 3. (a (b (c (d (c (f) ( (h 4. (a (b (c (d (c (f) ( (h same, lim X — > oo slower, lim X — » oo slower, lim x — > oo slower, lim X — » oo = lim ^-r- X — > oo e slower, lim X — > oo = ^0 = slower, lim X — > oo slower, lim X — » oo faster, lim - x — > oo lim k X — » oo 2 logio x lim 7r % x - X — > oo ( ln 10 ) e X — > OO On 10 ) e lim ,, ,L x X — ► OO ( ln 10)xe x 10x 4 + 30x+l lim X — » oo 40x 3 + 30 lim X — » oo xmj^x = Um e X — » oo = lim -^ = X — » OO xe" \/l+x 4 _ x(lnx- 1) lim X — > oo 120x 2 e x lim X — > 00 2 -Mix lim X — > 00 lim X — v oo ln x - 1 + 1 240 lim X — > oo lim 4# X — > oo e lim M X — ► oo 2e lim , X — > oo 4e 12x2 lim |% X — > OO 8e 2x 24 lim X — > 00 16e = lim (f x — > oo V2e ^ = lim 4r = X — > oo e lim x = oc X — > oo since £ < 1 slower, since for all reals we have — 1 < cos x < 1 < e c ' < e 1 => < lim <= X — ► OO ' same, lim X — » oo lim x-»oo so by the Sandwich Theorem we conclude that lim J X — > oo < % and also 5^ = lim X — > OO e ' (x-x+1) lim i X — > oo e same, lim X — ► oo faster, lim X — > oo same, lim X — > oo same, lim X — » oo x^ + 4x lim ^ti X — > OO 2x : lim (x 3 - 1) X — > oo v 7 lim X — > oo \/x 4 + > (x + 3) 2 lim X — > oo lim X — > oo :i + x) = v^i 1 = 1 lim X — > oo 2(x + 3) slower, lim X — > oo faster, lim \ X — > OO x 2 slower, lim — same, lim %- X — > OO x 2 same, lim X — > oo x ln x y2 lim X — > oo lim X — > oo 2x ln x x (ln 2) 2 X _ lim i X — » oo 2 2x : = lim 4 : x — ► oo e lim 8 = 8 X — > 00 lim X — > oo lim X — > oo lim l (In 2) 2 2" same, lim X — » oo slower, lim X — > oo slower, lim X — > oo 10x 2 = lim ( X — ► 00 v lim 10 = 10 x — > oo = lim i = X — > oo e x3/2/ log 10 x 2 X 2 3_ v 2 lim X — > oo . In 10 I f^ta lim In 10 x — > OO 2 In x x2 r4n lim In 10 x — > 00 (i) 2x rAn lim In 10 x — ► OO faster, lim x — > oo slower, lim X — > oo faster, lim ^^- = X — > OO x 2 same, lim x + j 00x X — > OO x 2 lim (x — 1) = oo X — » oo = x^oc «=° lim (•"'•'XiD x = n m 0" X — > OO 2x X — ► 00 .D 2 (i.i) x 2 = lim (l + 1Q0) = i x — y oo v x 1 Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 7.6 Relative Rates of Growth 455 5. (a) same, lim ^ lim X — > oo (Ml In x lim X — > oo 1 In 3 1 In 3 (b) same, lim ! j-^ = lim W X ^ OO ln x x ^ oo (£) (c) same, lim -p^ v ' X — > OO In x (d) faster, lim X — > oo (e) faster, lim X — > oo In. lim ^ X — > oo ln x lim k = k X — > OO 2 2 lim f = lim ffl£- X — > OO lnx X — > OO (1) lim tt^ = lim X — > OO 2yx x — » OO ^ CO lim X — > oo 1 lim x = oo X — > oo (f) same, lim ^^ = lim 5 = 5 w x ^ oo ln x x ^ oo (e) slower, lim -r^- = lim — f— V&/ ' X — > oo ln x x — > oo x ln x (h) faster, lim lim ff X — > OO (; lim xe x X — > oo 6. (a) same, lim ^^ = V ' X — > OO ln x X lim Mix 2 '. v'" 2 ) ln x 1 i: „ ln x 2 t—f; lim -j ln 2 x — > OO ln x r^ lim ^ ln 2 X — > OO ln x (b) same, lim X — » OO logio 10* ln x 1 n ]l)x *) In 10 I lim , X — > OO ln x -X_ ii m lnl0x In 10 x — » OO ln x In 10 x — ► OO (c) slower, lim / = lim OO (\A)(bix) r^ lim 2 ln 2 x — > OO 2 In 2 ,-Vn lim 1 = r^rs In 10 x — ► OO ™ ^ (d) slower, lim 4=^- V ' X — > OO ln x lim OO x2 In x (e) faster, lim x — > oo x— 2 ln x ln x lim X-tOO (r L --2) = ( lim r~) ~ 2 Unx I \ x _> oo lnxy lim x — > OO I I N -2 ( lim x ) \x — ► oo / (f) slower, lim £— V ' X — » oo ln x lim -^— X — > OO e ln x (g) slower, lim ^2^ V&/ X — > oo ln x lim X — > oo u*) lim ri- X — ► OO ln x (h) same, lim n , + v ' ' X — > OO In x lim X — > oo lim 2x bo 2x + 5 lim OO 2 lim 1 = 1 X — > oo 7. lim -§, X — > OO e ' = lim X — > oo lim e x / 2 = oo X — > oo e x grows faster than e x / 2 ; since for x > e e we have ln x > e and lim e X — > oo (ln x) x ' ln x \ x (In x) x grows faster than e x ; since x > ln x for all x > and lim lim X — > oo oo => x x grows faster than (ln x) x . Therefore, slowest to fastest are: e x / 2 , e x , (ln x) x , x x so the order is d, a, c, b lim X — > oo (In 2)" lim X — > oo (ln (ln 2))(ln 2)" 2x (In 2) x grows slower than x 2 ; lim lim x — ► oo V 2 (In(ln2)) 2 (ln2) x _ (ln(ln2)) 2 2 lim (ln 2) x = x — > oo 2x lim ..,,,. X — > OO ( ln 2)2 X X^OO 0n2) 2 2 x => x 2 grows slower than 2 X ; lim X — > oo lim X — > oo x — > oo e) = =^ 2 X grows slower than e x . Therefore, the slowest to the fastest is: (ln2) x ,x 2 ,2 x and e x so the order is c, b, a, d (a) false; lim - = v ' X — > oo x (b) false; lim -^ w X — > oo x + 5 (c) true; x < x + 5 (d) true; x < 2x => 2x j^= < 1 if x > 1 (or sufficiently large) < 1 if x > 1 (or sufficiently large) (e) true; lim -% X — » oo e (f) true: x + In x X ln x lim < 1 \-\ — 7= < 2 if x > 1 (or sufficiently large) Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 456 Chapter 7 Transcendental Functions (g) false; lim fe = Inn ln2x ,^4- = lim 1 = 1 (h) true; ^p^ < ^ {x + 5)2 < >i±5 = i + | < 6 if x > 1 (or sufficiently large) 10. (a) true; /{\ = ^3-3 < 1 if x > 1 (or sufficiently large) w (b) true; 44 = 1 + j < 2 if x > 1 (or sufficiently large) (c) false; lim v> i 2 J x -» 00 (l) (d) true; 2 + cos x < 3 - : °" x 3 (e) true; ^ lim (1 - i) = 1 x — > 00 v xy < ^ if x is sufficiently large 1 + 4 and 4 — > Oasx — > 00 =>• l + 4<2ifxis sufficiently large (f) true; lim ^ = lim ^ = lim -^ X ^ OO x x^oo x x — > 00 1 (g) true; -221) < |f = 1 if x is sufficiently large (h) false; lim lim X — > OO ln(x 2 +l) x _> co (2* lim 4^ X — > cx> 2x2 lim (i + =ij) = ± X — > (X) v 2 2x y 2 11. If f(x) and g(x) grow at the same rate, then lim 44=1-7^0 => lim v ' 6V y 6 x — > 00 sW ~ x — > c M _ T a. n - 1™ SW - . I ^ 0. Then OO « x > L L < 1 if x is sufficiently large =>L-1<-^<L+1 => ^ < |L| + 1 if x is sufficiently large |g(X) =>• f = 0(g). Similarly, jg < |±| + 1 => g = 0(f). fix) 12. When the degree of f is less than the degree of g since in that case lim 44 = 0. b b b x _> qq g( x ) fix) 13. When the degree of f is less than or equal to the degree of g since lim 44 = when the degree of f is smaller b 1 fc & x _> oq g(x) b f(x) than the degree of g, and lim 44 = § (the ratio of the leading coefficients) when the degrees are the same. 14. Polynomials of a greater degree grow at a greater rate than polynomials of a lesser degree. Polynomials of the same degree grow at the same rate. 15. lim x — > 00 ln(x+l) In x x+1 , lim x — >■ 00 (i lim -iv = lim i = 1 and lim '"< x + " 9) x— > 00 x+1 x^oo 1 x— > 00 lnx lim x — > 00 X + 999 J lim , x „ nn X — » OO x + 999 16. lim lr 4°^ x — > 00 i |lx lim X — > CO . x + a lim -f- X — > CO x + a lim 1 X-tOO ' 1. Therefore, the relative rates are the same. 17. lim x — > 00 4iQx+i lim X — > CO 10x+l 10 and lim x — » 00 /x+l lim =ti = v 1 = 1. Since the growth rate x — > 00 x v 6 is transitive, we conclude that x/lOx + 1 and \/x + 1 have the same growth rate (that of x/x) 4 x4 + x _ 18. lim x — *• 00 x ' lim *-+* = 1 a nd lim ^f^ X — > OO x X — » 00 x lim - / x — > 00 x 1 . Since the growth rate is transitive, we conclude that \J x 4 + x and y x 4 — x 3 have the same growth rate (that of x 2 ) . 19. lim 4 x — > 00 e lim x — > 00 lim 4=0 X — > CO e x n = o (e x ) for any non-negative integer n Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.6 Relative Rates of Growth 457 20. Ifp(x) = a n x n + a n _ix n aix + an, then lim u X — > oo p(x] a n lim ^r + a n _ i lim x — > OO e X — *• oo p(x) ai lim 4 + an lim 4 where each limit is zero (from Exercise 19). Therefore, lim - 1 -— = x ^ co e u x ^ co e x ^ co $■ e x grows faster than any polynomial. 21. (a) lim ^ = lim s^ v/ X^OO lnx X — > OO n(i) i) lim x 1/n n ' X — > CO 1/10 6 (b) m (e 17 > 000 < 000 ) = 17,000,000 < (e 17xl ° 6 ) (c) x« 3.430631121 x 10 15 (d) In the interval [3.41 x 10 15 ,3.45 x 10 15 ] we have In x = 10 In (In x). The graphs cross at about 3.4306311 x 10 15 . IT co =4> In x = o (x 1 ' n ) for any positive integer n i 24,154,952.75 0.004' y = In x- 10 ln(ln x) 0.002 -0.002 3.41 io y^ 3.45 10 -0.004 yS -0.006 / 22. lim X — > CO lim In x lim af noo V x ) a„x" + a„_, x n_l + . . . + ajx + ag (■!) lim x — > oo M ( nx ") => In x grows slower than any non-constant polynomial (n > 1) 23. (a) lim n log2 n CO n (logi n) 2 lim OO lo E2 n => n log2 n grow$>) slower than n (log? n) ; lim ° n — > oo n log2 n n3/2 I — ] hm -^7/- n — > co n ' r^TT lim In 2 n , W _ _2_ lim J_ CO (i) n-V2 - In 2 n ™CO "1/2 =4> n log2 n grows slower than n 3,/2 . Therefore, n log2 n grows at the slowest rate => the algorithm that takes 0(n log2 n) steps is the most efficient in the long run. 24. (a) lim n — > co (log, n) 2 lim n — > co lim ^% n — > oo n < ln 2 r ,. 2 < lnn )(n) lim ,. .A 7 n — » oo ( ln 2 ) 2 lim lim (In 2)2 n^+ oo i (ln 2) 2 n _> go n =>• (log2 n) 2 grows slower than n; lim ^2ip>L = lim ^ n — > oo v n lo S2 n n — > oo v n I — 1 i • V in 2 ; hm ^ n — > CO n 1/2 r^> lim ln 2 n — > oo n In n 1 2 (b) 20 40 60 80 100 = r^ lim , n v " ,,, = r4; lim -k= = =>- (log2 n) grows slower than \/n log2 n. Therefore (log2 n) grows In 2 x — ► OO (2) n ' ln 2 n — > CO 1 ' v at the slowest rate => the algorithm that takes O ((log2 n) 2 ) steps is the most efficient in the long run. 25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because 2 19 = 524,288 < 1,000,000 < 1,048,576 = 2 20 . 26. It could take 450,000 steps for a sequential search, but at most 19 steps for a binary search because 2 18 = 262,144 < 450,000 524,288 2 19 . Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 458 Chapter 7 Transcendental Functions 7.7 INVERSE TRIGONOMETRIC FUNCTIONS 1. (a) (b) (c) 3- (a) -| (b) f (c) -f 5. (a) f (b) 3 f (c) 1 7. (a) 3 f (b) f (c) 2 f 9. (a) I (b) -\ (c) | 11. (a) 3tt (b) I (c) 2tt 2. (a) 4. (a) | 6. (a) & 8. (a) | 10. (a) -f 12. (a) 5 (b) \ (b) (b) I (b) (b) | (b) 5- (C) (c) I (0 f (c) f (c) f 13. a = sin : (-^) =4> cos a = ||, tan a = ^, sec a = ||, esc a = y, and cot a = y 14. a = tan -1 (|) =>■ sin a = I, cos a = 1, sec a = |, esc a = |, and cot a = | sin a = -75, cos a = \-, tan a = —2, esc a = ^-, and cot a sin a , cos a j=, tan a = — |, esc a = ^— , and cot a = — | 18. sec (cos" 1 I) =sec(f) =2 15. a = sec : 16. a = sec" 1 (- ^) 17. sin [cos- 1 &\ = sin (f) = 4- - 1 (-I))=tan(-I) = -^ 20. cot(sin- 1 (-f))=cot(-f) sec -1 2) + cos (tan- 1 ( — \J 3] j = esc (cos -1 (|)) + cos sec" 1 1) + sin (esc- 1 (-2)) = tan (cos" 1 \) + sin (sin" 1 (-5)) = tan(0) + sin (- |) =0+ (- |) = - | 19. tan 21. esc 22. tan 23. sin 24. cot 1 v/3 |)=csc(f) + cos(-f) = ^ + i = ^# sin- 1 (-I)+cos- 1 (-i))=sin(-| + f) sin(f) = l sin" 1 (- I) - sec" 1 2) = cot (- f - cos" 1 (±)) = cot (- f - i) =«*(-!) =0 25. sec (tan -1 1 + esc -1 1) = sec (f + sin -1 \) = sec (f + |) = sec 26. sec (cor 1 ^3 + esc" 1 (-1)] = sec (f + sin" 1 (i)) 27. sec" 1 (sec (- |)) = sec" 1 (-4) - — - 1 ^ 28. cor 1 (cot (- I)) = cor 1 (-1) = : 3 f) TT_ TT_ 7T1 2 3 2) -V~2 ?2E 4 Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.7 Inverse Trigonometric Functions 459 29. a = tan l | indicates the diagram sec (tan" 1 |) = sec a = V^±4 30. a = tan : 2x indicates the diagram 2x V4x 2 + 1/ v^o 1 sec (tan : 2x) = sec a = y 4x 2 + 1 31. a = sec : 3y indicates the diagram y] 9 f-i => tan (sec : 3y) = tan a = y^y 2 - 1 32. a = sec l I indicates the diagram tan (sec 1 i J = tan a = ^-j — -l y\ 33. a = sin x x indicates the diagram » =>- cos (sin : x) = cos a = y 1 — x 2 34. a = cos x indicates the diagram 35. a = tan J v x 2 — 2x indicates the diagram V x 2 — 2x Vi-x 2 =>• tan (cos 1 x) = tan a \/l-x 2 x' i -2x =^ sin (tan" 1 \/x 2 - 2x) sin a x- 1 36. a = tan : , * indicates the diagram sin ( tan 1 , | ) = sin . \ \/ X 2 + 1 / \/2x 2 + 1 37. a = sin : -y indicates the diagram =4> cos (sin : -j) = cos a = 3 y 38. a = sin 1 | indicates the diagram =>• cos (sin : I) = cos a y/25^7 39. a = sec x | indicates the diagram =>- sin sec 4 = sin a 1 x\ - cln™ - \/x 2 -16 Cf|t (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 460 Chapter 7 Transcendental Functions 40. a = sec 1 ^ x + indicates the diagram 2 Vx 2 + 4/ S^ix x ■ ( -1 \/x 2 + 4\ sin I sec ^— ^ — J = sin . V^x 2 + 4 41. lim sin x x^ 1- 42. lim cos : x = 7r x->-l+ 43. lim tan _1 x = g x — > oo 2 44. lim tan -1 x X — > — oo 45. lim sec x = 5 X — > oo l 46. lim sec : x= lim cos : (-) = % X — > — OO X — > — oo \^> * 47. lim esc : x= lim sin : (-) =0 48. lim esc J x= lim sin l (-) =0 X — > — 00 X — » — 00 v "/ 49. y = cos" 1 (x 2 ) => | 2\ -2x /l - (x 2 ) 2 Vl^ 50. y = cos l {-) = sec : x dy _ 1 dx |x| Vx 2 -1 51. y = sin-V2t => f - ------ 1 - v/2t 2 Vl-2t 2 52. y = sin- 1 (1 - t) => S dy dt x/l-(l-t) 2 VSt^? 53. y = sec" 1 (2s + 1) Jv |2s+l| v^s+l) 2 -! |2s+l| x/4s 2 + 4s |2s+l| x/s 2 + s 54. y = sec 1 5s => dy |5s| x/(5s) 2 -l s| x/25s 2 - 1 55. y = esc- 1 (x 2 + !)=>.£ 2x -2x dx |x 2 +l| \/(x 2 +l) 2 -l (x 2 +l)v'x I T 2\- dy dx 56. y = esc : (|) 57. y = sec -1 (1) = cos -1 1 l!lv / (l? rT wV^ 4 ' |x|v/ *^ dy dt / t _ t 2 58. y = sin t 2 J CSC -(«) dy dt |.2| ./A? /t£-9 tvt 4 — 9 59. y = cor 1 x/t = cor 1 1 1 / 2 => f iK^ l + (tV 2 ) 2 2x/t(! +0 60. y = cor 1 x/t- 1 = cor 1 (t - l) 1 / 2 =*► g = - Q)<'-')- 1/2 = -i = -i l + [(t-l)!/ 2 ] 2 2^/1^1(1+1-1) 2tx/t--T 61. y = In (tan l x) dy _ li+x^J dx tan -1 x (tan-lx)(l+x 2 62. y = tan -1 (In x) dy = (j) = l dx l+(lnx) 2 x[l+(lnx) 2 ] 63. y = esc- 1 (e') => g UA e'U/e'V-l x/e 2 ' - 1 Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.7 Inverse Trigonometric Functions 461 64. y = cos x (e ') => dy dt v/l-(e-) 2 Vl-e" 2 ' 65. y = svT^ S 2 + cos : s s 2 = 8(1 \/l-s 1 Vl-s 2 - sec -1 s = l/l-S 2 = Vs 2 - s |s|-l 1 = (s 2 - ,1/2 2\l/2 .2N-1/2 s (1 - s 2 ) v " + cos- 1 s => | = (1 - s 2 ) v/ + s (|) (1 - s 2 r v V2s) - V 7 ! - s 2 - s 2 + 1 _ 1 - s 2 - s 2 - 1 _ -2s 2 'l-s 2 x/l-s 2 M-s 2 s| Vs 2 - 1 Vs 2 - 1 |s| Vs 2 - 1 |s| Vs 2 -1 67. y = tan- i yx 2 ^T + csc- 1 X = tan- 1 (x 2 -l) 1/2 + csc- 1 X =► g- iA 1 '^' ' "^ 1+ ( x 2 -!) 1 ' 2 ] 2 Wv 7 ^ 3 ! x v x 2 — 1 |x| v x 2 — 1 0, for x > 1 68. y = cor 1 (!) - tan' 1 x = f - tarr 1 (x" 1 ) - tan" 1 x => ^ = =^U - -^ ' U/ 2 v / dx l + (x _1 ) 1 + _ _J L_ - n -x 2 ~~ x 2 +l 1+x 2 — U 69. y = x sin x x + ^1-x 2 = x sin- 1 x + (1 - x 2 ) 1/2 => | = sin" 1 x + x (-7^) + (5) (1 - x 2 ) _1/2 (-2x) \7i-x 2 \/i~ 70. y = ln(x 2 + 4)-xtan- 1 (|) => | = ^ - tan" 1 (f ) - x i+(ir 2x x 2 + 4 tan -1 (x\ _ 2x \2) 4 + x 2 71. ^-tan- 1 ^) /75b dx = sin_1 (f)+C 72. f . x dx = i f . 2 , dx = i f^- , where u = 2x and du = 2 dx J V1-4X 2 2 J Vl-(2x) 2 2 J ^l- u 2 ! sin' 1 u + C = ! sin" 1 (2x) + C 73- J TT ^dx=J (v/n) +x 2 dx 4- tan" 1 -£- + C 74. /^ dx = |/p^ dx = ^ tan- 1 (^) + C = f tan" 1 (^) + C 75. f , d * = f , d " , where u = 5x and du = 5 dx J xV25x 2 -2 J u\/u 2 -2 = 73 sec "l73| +c= 75 sec "l^l +c 76. f /* = f , d " , where u = J~5x and du = a/5 dx J x\/5x 2 -4 J uVu 2 -4 v v = isec- 1 |f|+C= isec- 1 |^P|+ C 11 ■ II 707, = [ 4 sin_1 f] = 4 ( sin_1 I - sin_1 °) = 4 (I - °) Copyright (c) 1 Pearson Etati, Inc., publishing as Pearson Addison-Wesle 462 Chapter 7 Transcendental Functions ,3\/2/4 78 r 2/4 ^ = i r Jo V9 - 4s 2 2 Jo ., , , du „ , where u = 2s and du = 2 ds; s = 2 Jo \/9-u 2 [I sin- |] f /2 = I (rin-'f - sin- o) = J (} - 0) = f n 3v/2 u = 0, s = ~y 3^2 '2 dt; t = => u = 0, t = 2 ^> u = 2\/ 2 79 - ThtW = jify&?> where u = ^ and du ~ J - 1 (tan- 2j| - tan- o) = i (tan" 1 1 - tan" 1 0) = i (| - 0) = $ 4s tan" 1 -jj- , 80 - Il 2 4+W = 75 f-zJi 4+^P ' where u = V3t and du = v^ dt; t = -2 => u = -2a/3, t = 2 => u = 2^3 81./. -V2/2 dy -V5 2 p-\/2 , I — , " , where u = 2y and du = 2 dy ; y = — 1 =>• u= —2, y J -2 u\/u 2 ' [sec- 1 |u|] _a 2 = sec" 1 -a/2 - sec" 1 |-2| = f - f => u= -V2 -v^ 82. f /n dy 2 ; = f /" , where u = 3y and du = 3 dy; y = - \ =» u = -2, y J -2/3 yx /9y 2 -l J -2 u\/u 2 -l ^ •"•' 3 > ■> -2/3 yV^y 2 - 1 J-i u\/u 2 -l [sec- 1 |u|]I 9 v/5 = sec" 1 l-y^l - sec" 1 |-2| 4 3 VI ■y/2 83. / 3 dr ,/l -4(r-l)2 2 J /T3 = | sin -1 u + C = 2 f , du , where u = 2(r - 1) and du = 2 dr J \/l-u 2 v ' 3 sin- 1 u + C = | sin" 1 2(r - 1) + C 84- / v ^ T7 =6/ v ^,whereu = r+landdu = dr 6 sin -1 | + C = 6 sin" 85. f , dx n2 = f ^S , where u = x - 1 and du = dx J 2 + (x- l) 2 J 2 + u 2 = 72 tan "73 + c =73 tan "(7T) +c 86 - / i + (3 d x+i) 2 = 3 JlTTP - where u = 3x + 1 and du = 3 dx = i tan" 1 u + C = i tan" 1 (3x + 1) + C 87. / (2x-l) v /(2x-l) 2 -4 i I — 4s — , where u = 2x — 1 and du = 2 dx 1 J uvu 2 - 4 ! - ! — " l |u| ' C = isec- 1 1^1 h J u\/u 2 (x + 3)v'(x + 3) 2 -25 J u\/u 2 -25 i _„-l Iu| , c= l sec -i v_: +c where u = x + 3 and du = dx I sec- 1 |^| 89. r 2 rxr^il = 2 f ' -rrS . where u = sin 9 and du = cos 9 d0; 6> = - ? J _,r/2 1 + (sin 9) 2 J_i 1+u 2 ' ' 2 = ptan-u]^ =2 (tan- 1 1 -tan^C-l)) = 2 [f - (- |)] = tt =^u= -1, |^u=l Cfigl (c) 1 Pea Education, Inc., publishing as Pearson Addison-Wesle Section 7.7 Inverse Trigonometric Functions 463 90 - JT/6 Tf^ = -/^TT^' whereu = cotxanddu =- csc2xdx ; x =I = > u= v / 3,x=| :> u=l = [- tan -1 u] ] ^5 = - tan -1 1 + tan -1 \pb ■ 7T _i_ 7T _7T_ 4 ' 3 ~~ 12 91. I Jo e* dx - | T^-o , where u = e x and du = e x dx; x = => u=l,x = ln v3 =^ u = v3 1 + e?" J i 1 +U- 2 ' ' ' v v = [tan- 1 u] /* = tan" 1 a/3 - tan" 1 1 = § - * = jl 92. f , n lf 2 ^ = 4 r t^ , where u = In t and du = ± dt; t = 1 => u = 0, t = e^/ 4 => u = f J I t ( 1 + m z t) J o 1 + u t 4 = [4 tan" 1 u] q /4 = 4 (tan" 1 | - tan" 1 0) = 4 tan" 1 f 93. f ^hr = 5 f^— . where u = y 2 and du = 2y dy •J \/i - y 4 2j v /|_ u 2 = | sin -1 u + C = \ sin -1 y 2 + C 94. / / , du , where u = tan y and du = sec 2 y dy sec 2 y dy v/l-tan 2 y ~~ J /T = sin -1 u + C = sin -1 (tan y) + C J \/-x 2 + 4x-3 J dx ./-x 2 + 4x-3 J N /l-(x 2 -4x + 4) J x /l-(x-2)2 J: dx sin -1 (x - 2) + C 96. f ^2= = f J \/2x-x 2 J dx x /l-(x 2 -2x+l) J Vl-(x-l)- J- dx sin" 1 (x - 1) + C 97 - £ ya=? = 6 I V (*+»+.> = 6 L 7 w J ww = 6 [-*- 1 m] -i V3-21-1 2 J-i v / 4 -( t2 + 2t + 1 ) "J-ix/2 2 " 6 [sin" 1 (i) - sin" 1 0] = 6 (| - 0) = tt I, ' 6dt _ -x f ' 2dt Jl/ /2 ^/3 + 4t-4t 2 Ji/2 % /4-(4t 2 -4t+l) J 1/2 % /2 2 -(2t-l) 2 3 r 1 - J 1/2 • 2dt 3 [-- 1 (3^)] I 3 [sin l (i\ ^0] =3(f-0) = f /2 on P dy _ P dy _ P "• J y 2 -2y + 5 ~ J 4 + y 2 -2y+l — J 1 , , d y _ I tan -i (y^A\ i r y 2 -2y + 5 ~~ J 4 + y 2 -2y+l — J 2 2 + (y-l) 2 ~~ 2 Um I 2 J ^ ^ loo- /.^no = / dy y 2 + 6y+10 J l + (y 2 + 6y + 9) J l + (y + 3) 2 J, dy tan" 1 (y + 3) + C ioi. r^fn = 8 /, 2 x 2 -2x + 2 - U J, !+(x 2 -2x+l) - 8 J, l+(x-l) 2 - 8 t tan 1( - X ~ ! )ll 8 (tan- 1 1 - tan" 1 0) = 8 (f - 0) = 2tt i° 2 - r^^o= 2 r iK -' ,i iK € 2 x 2 -6x+10 J 2 l + (x 2 -6x + 9) J 2 l+(x-3) 2 = 2 [tan- 1 1 - tan- 1 (-1)] = 2 [J - (- f )] = tt 2 [tan- 1 (x- 3)] ^ 103 f ^ = f ^ r J (x+l)Vx 2 + 2x J (x+l)Vx 2 + 2x+l-l J ( ;+l)Vx 2 + 2x J (x + 1) Vx 2 + 2x + 1 - 1 J (x+ l^x + l) 2 - 1 f — , d " , where u = x + 1 and du = dx J UA/U 2 — 1 sec -1 lul +C = sec -1 |x + 1| + C Copyright (c| 1 Pea Education, k, publishing as Pearson Addison-Wesle 464 Chapter 7 Transcendental Functions dx I dx J (x- — =f- i + 4 - 1 J (x - J (x-2)\/x 2 -4x + 3 J (x-2)Vx 2 -4x + 4-l J (x-2)\/(x- 2) 2 - 1 = / — ,\ du, where u = x — 2 and du = dx J uy u 2 — 1 sec" 1 |u| + C = sec" 1 |x - 21 + C 105. I e r" dx = I e" du, where u = sin 1 x and du \/l-x 2 e u + C = e sin X + C 106. | e , c °" dx = — e" du, where u = cos l x and du -dx yr -e" + C = -e cos x + C 107. I ^S — — dx = | u 2 du, where u = sin 1 x and du J VI -x 2 J dx V^ 108. J ^+J" dx = Ju 1 / 2 du, where u = tan" 1 x and du dx 1 + x 2 | u 3/2 + C = | (tan" 1 x) 3/2 + C = § ^/(tan- 1 x) 3 + C 109 - J i5fwtt) d y = / W dy = /u du ' where u = tan_1 y = ln|u|+C = ln|tan- 1 y|+C and du = ^-j 1+y 2 110. / dy = I sin-^ ^y = I u ^ u ' wnere u = sin 1 y and du (sin- 1 y) Vl+y 2 = In |u| +C = ln|sin- 1 y| +C dy in. r' sec2 ' se ;" ix) dx= r /3 s Ji/2 x\/x 2 -l Jir/4 \/2 x\/x 2 - 1 Jir/ 4 [tan u]^ 3 = tan §- tan f = ^ - 1 sec 2 u du, where u = sec x x and du = — , dx ; x = \/2 =>• \/x 2 - 1 u=f,x = 2 1 12. / _ cos ^ — ^ dx = I cos u du, where u = sec 1 x and du = — 4 s — ; x = —, J 2 A/1 x\/x 2 -l Jtt/6 x\/x 2 -1 J u=g,x = 2 [sinu]^ 6 = sin|-sin| = ^— 113. lim Msziix = lim Wi-25x2; =5 x^O x x^O ' 1 14. lim n/^i - i; m M^i lim 2 _ , M* x^l+ scu x x-*!-* lim x^ 1+ 1) (x 2 -l)- 1 / 2 (2x) 115. lim x tan X — > CO -1 (T\ lim X — > oo tan" 1 (2X" 1 ) 116. lim x^0 2 tan" 1 3x 2 7x 2 • r i ^ I x| v'x 2 -l / C -2x-2 \ lim X — > CO Vi+4x~ 2 y -X" 2 6 6 lim x |x| = 1 x — > l + lim . , . _ 2 x — > CO' l+4x~ 2 lim ,,. = lim , . x ^0 14x x^0 7(1+9x4) 7 Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.7 Inverse Trigonometric Functions 465 117. Ify = lnx- | ln(l+x 2 ) - ^_p + C, then dy X 1 +x 2 dx — I I _ x _ 1 , tan" 1 x \ ~~ \x 1+X 2 X(l+X 2 ) + X 2 J which verifies the formula H V _ xtl+x^-x^x+ttan^xJtl+x 2 ) , _ tan" 1 x Hy ax- x 2(i + x2) QX - x 2 QX ' 118. Ify=^cos-5x + |/ 7 ^dx,thendy=[x3cos-5x + (^)( 7r ^)+|( 7r ^)]dx (x 3 cos : 5x) dx, which verifies the formula 119. If y = x (sin- 1 x) 2 - 2x + 2yJ \ - x 2 sin" 1 x + C, then j [/ ■ _i \2 2x (sin 1 x) /■, dy = [(sin x) + v ) i _ x2 i - 2 the formula + -^ sin" 1 x + 2^1 -x 2 ( -r±— )] dx = (sin" 1 x) 2 V 1 — x 2 v V i — * / J dx, which verifies ln(a 2 + x 2 ) + p 2 ^ I -2 ■ : 120. Ify = xln(a 2 +x 2 ) -2x + 2atan~ 1 (|) + C, then dy = [in (a 2 + x 2 ) + 2 (pipfs) -2\ dx = In (a 2 + x 2 ) dx, which verifies the formula 1+5 dx 121. dy _ 1 dx yr: dy dx yr =>. y = sin : x + C; x = and y = => = sin x + C => C = =>- y = sin -1 x dy 122. ~g = jjL - 1 =>• dy = (y^ - l) dx =>• y = tan" 1 (x) - x + C; x = and y = 1 => 1 = tan" 1 - + C =>■ C = 1 => y = tan -1 (x) - x + 1 123. ^ dx XV x 2 — 1 dy =- ,"* =» y = sec -1 |x| + C; x = 2 and y = 7r => ir = sec -1 2 + C => C = 7r - sec -1 2 d" x\/x 2 - 1 = 7r-f = f => y = sec" 1 (x) + f , x > 1 124. dy _ _J 2 dx 1 + x 2 yjT d y=(ir^ v^ dx =>■ y = tan : x — 2 sin x x + C; x = and y = 2 => 2 = tan" 1 0-2 sin" 1 + C => C = 2 =>• y = tan" 1 x - 2 sin" 1 x + 2 125. The angle a is the large angle between the wall and the right end of the blackboard minus the small angle 126. V = tt £ 3 [2 2 - (sec y) 2 ] dy = tt [4y - tan y] q /3 = tt (% - v^) 127. V = (i) 7rr 2 h = Q) tt(3 sin 6>) 2 (3 cos 0) = 9tt (cos - cos 3 0), where < 9 < S -v d_V 2 ^ dfl -97r(sin(9)(l -3 cos 2 =>• sin = or cos = ± -4- =>■ the critical points are: 0, cos l ( -4- j , and cos : ( \- j ; but 1 I j- ] is not in the domain. When = 0, we have a minimum and when 6* = cos" 1 ( -4- j w 54.7°, we cos I — have a maximum volume 128. 65° + (90° -P) + (90° - a) = 180° => a = 65° - (3 = 65° - tan" 1 (|) w 65° - 22.78° « 42.22° 129. Take each square as a unit square. From the diagram we have the following: the smallest angle a has a tangent of 1 =>• a = tan" 1 1; the middle angle (3 has a tangent of 2 =>• /? = tan" 1 2; and the largest angle 7 has a tangent of 3 =>• 7 = tan" 1 3. The sum of these three angles isw =4> a + (3 + j = tt =>- tan" 1 1 + tan" 1 2 + tan" 1 3 = tt. Copyright [c| 1 Pearson Elation, In, publishing as Pearson Addison-Wesle 466 Chapter 7 Transcendental Functions 130. (a) From the symmetry of the diagram, we see that tt — sec -1 x is the vertical distance from the graph of y = sec -1 x to the line y = 7r and this distance is the same as the height of y = sec -1 x above the x-axis at — x; i.e., 7r — sec -1 x = sec -1 (— x). (b) cos -1 (— x) = 7r — cos -1 x, where — 1 < x < 1 =>• cos -1 (— -) = n — cos -1 (-), where x > 1 or x < —1 =>• sec -1 (— x) = 7r — sec -1 x 131. sin -1 (1) + cos -1 (1) = f + 0= f ; sin" 1 (0) + cos" 1 (0) = + § = | ; and sin" 1 (-1) + cos" 1 (-1) = - § + tt = f. If x 6 (—1,0) and x = —a, then sin -1 (x) + cos -1 (x) = sin' 1 (—a) + cos -1 (—a) = — sin -1 a + (tt — cos -1 a) = tt — (sin -1 a + cos -1 a) = tt — | = f from Equations (3) and (4) in the text. 132. x => tan a = x and tan (3 a + /3 tan 1 x + tan : ; 133. (a) Defined; there is an angle whose tangent is 2. (b) Not defined; there is no angle whose cosine is 2. 134. (a) Not defined; there is no angle whose cosecant is \. (b) Defined; there is an angle whose cosecant is 2. 135. (a) Not defined; there is no angle whose secant is 0. (b) Not defined; there is no angle whose sine is y 2. 136. (a) Defined; there is an angle whose cotangent is — | . (b) Not defined; there is no angle whose cosine is —5. 137. a(x) = cor 1 (§) - cor 1 (|) , x > =* a'(x) = ^ + ^h = ""(WH^)"^ ; solvin S a'(x) = => -135 - 15x 2 + 675 + 3x 2 = =4> x = 3a/5 ; a'(x) > when < x < 3^ and a'(x) < for x > 3 v 5 =>• there is a maximum at 3y5 ft from the front of the room 138. From the accompanying figure, a + j3 + 6 = tt, cot a = y and cot /? = ^ =4> 9 = tt - cor 1 x - cor 1 (2 - x) 1 a \ B / 'e X >. de _ _\ i _ i+(2-x) 2 -(i+x 2 ) dx 1+x 2 l+(2-x) 2 (l+x 2 )[l+(2-x) 2 ] = (i + x 2 )V + 4 (2-x) 2 ] solving d £=0 => x = l;g > Ofor < x < 1 and f <0forx>l =>• at x = 1 there is a maximum 8 = tt — cot -1 1 — cor 1 (2— 1) = tt — f — § = f 139. Yes, sin J x and —cos : x differ by the constant | 140. Yes, the derivatives of y = — cos x x + C and y = cos : (— x) + C are both —A v'l-x 2 141. esc -1 u = f - sec -1 u => £ (csr 1 u) = £ (| - sec" 1 u) = j ii |u| v u 2 — 1 |u| v u 2 — 1 ,H>i Cf|t (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.7 Inverse Trigonometric Functions 467 142. y = tan : x =>• tan y = x => £(tany)=£(x) (sec 2 y) dx 1 dy _ 1 dx sec 2 y (^ 1+x 2 1+X 2 , as indicated by the triangle 143. f(x) = sec x =4> f'(x) = sec x tan x dt- 1 dx dx lx = f-l(b) secfsec- 1 b)tan(sec-! b) b(±\/b 2 -l) ' Since the slope of sec x is always positive, we the right sign by writing x-sec x xl^x 2 -!' 144. cot -1 u = | - tan -1 u =>• j- (cot -1 u) £( f _tan- 1 u)=0- T f^ 1+u 2 145. The functions f and g have the same derivative (for x > 0), namely r } . The functions therefore differ v x ' x "*" ' by a constant. To identify the constant we can set x equal to in the equation f(x) = g(x) + C, obtaining sin -1 (-1) = 2 tan" 1 (0) + C => |=0 + C =>■ C = - f . For x > 0, we have sin" 1 (|=i) = 2 tan" 1 \fc-\ 146. The functions f and g have the same derivative for x > 0, namely y— ^ • The functions therefore differ by a constant for x > 0. To identify the constant we can set x equal to 1 in the equation f(x) = g(x) + C, obtaining - 1 (AS\ = tan" 1 1 + C =4> | = f + C =^> C = 0. For x > 0, we have sin" l l sfx 2 + I tan -l l *vs *vs 147- V = tt / ^ (^i-) dx = n J_^ /3 ^ dx = tt [tan- 1 x]_^ /3 = tt [tan" 1 ^3 - tan" 1 (- f )] = *[f-(-!)]=T 148. y = VT3rf = (l-x 8 ) 1/2 ^ y ' = (i) ( i_ x 2 ) -i/2 ( _ 2x ) => l + (y') 2 = T ^;L=/^l + (y') 2 dx = 2 X' /2 Tib dx = 2 I siirl ^ = 2 (1 " °) = I 149. (a) A(x) = | (diameter) 2 = f [-^ - = ^[tan- 1 x]i 1 = (^)(2)(|) = f (b) A(x) = (edge) 2 = [-^i- - (- ^_) v/T^ 1+x 2 V=£A(x)dx=£^ dx^ + X 2 1+X 2 v=/;A(x)dx=/_; 4dx 1+x 2 4[tan" 1 x]i 1 = 4[tan" 1 (l)-tan" 1 (-l)] = 4 [f - (- f )] = 2tt 150. (a) A(x) = g (diameter) 2 = J (^ - o)* = \ [^-^ \/l-x 2 V £ A « dx rX 7^ dx = * [sin " x] -^ = * K 1 (^) - sin " (-#)]=- tf - (- 1)] = t (b) A(x ) =«i T ^ = i(^_ )- = 7 2[sin" 1 x]! 1 ,1 fi/ 2 V*/2 2(?-2}=7T l-x 2 V = X A(x) dx = J ■fin- \J\--i? dx 151. (a) sec" 1 1.5 = cos" 1 ^ » 0.84107 (c) cot" 1 2 = f - tan" 1 2 w 0.46365 152. (a) sec" 1 (-3) = cos" 1 (- \ 1.91063 (c) cor 1 (-2) = f - tan" 1 (-2) rj 2.67795 (b) esc" 1 (-1.5) = sin" 1 (b) esc : 1.7 = sin 1.5/ -0.72973 0.62887 Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 468 Chapter 7 Transcendental Functions 153. (a) Domain: all real numbers except those having the form | + kn where k is an integer. Range: - § < y < | (b) Domain: — oo < x < oo; Range: — oo < y < oo The graph of y = tan' 1 (tan x) is periodic, the graph of y = tan (tan -1 x) = x for — oo < x < oo. y y = tan" (tan x 3it ■3k -k. -3it y = tan (tan-'x) k 3n 154. (a) Domain: — oo < x < oo; Range: — | < y < | (b) Domain: —1 < x < 1; Range: — 1 < y < 1 The graph of y = sin -1 (sin x) is periodic; the graph of y = sin (sin -1 x) = x for —1 < x < 1. 71 ~2 y = sin (sin x) / \ 2 -2* -22. 2 -it 2. it\ /zk 2 \S -2 y = sin (sin- 1 x) 1 2 155. (a) Domain: — oo < x < oo; Range: < y < ir (b) Domain: —1 < x < 1; Range: — 1 < y < 1 The graph of y = cos -1 (cos x) is periodic; the graph of y = cos (cos -1 x) = x for — 1 < x < 1. y = cos" (cos x ) y = cos \cos~ xj Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesle Section 7.7 Inverse Trigonometric Functions 469 156. Since the domain of sec -1 x is (— oo, — 1] U [1, oo), we have sec (sec -1 x) = x for |x| > 1. The graph of y = sec (sec -1 x) is the line y = x with the open line segment from (—1, —1) to (1, 1) removed. c(sec x) 157. The graphs are identical for y = 2 sin (2 tan l x) = 4 [sin (tan- 1 x)] [cos (tan" 1 x)] = 4 (-^) (-^) ^777 = ^^j from the triangle y = 2 sin (2 tan -1 x) . 158. The graphs are identical for y = cos (2 sec l x) = cos 2 (sec -1 x) — sin 2 (sec -1 x) = \ — ^^ 2-x -j- from the triangle /7TT y 5C 4C 3C y = cos(2sec~ 1 x) / \ 2-x 2 '" x 2 "tO =3 — s » — 159. The values of f increase over the interval [— 1, 1] because f ' > 0, and the graph of f steepens as the values of f ' increase towards the ends of the interval. The graph of f is concave down to the left of the origin where f" < 0, and concave up to the right of the origin where f " > 0. There is an inflection point at x = where f " = and f ' has a local minimum value. 160. The values of f increase throughout the interval (—00, 00) because f ' > 0, and they increase most rapidly near the origin where the values off are relatively large. The graph of f is concave up to the left of the origin where f " > 0, and concave down to the right of the origin where f " < 0. There is an inflection point at x = where f " = and f has a local maximum value. Copyright (c| 1 Pea Education, k, publishing as Pearson Addison-Wesle 470 Chapter 7 Transcendental Functions 7.8 HYPERBOLIC FUNCTIONS 1 . sinh x : coth x | => cosh x = \f\ + sinh 2 x = J 1 + (- |) 9_ 16 i = ! , tanh x = ^i 16 4 cosh x tanh x sech x cosh x I , and csch x = -J— = — t 5 ' sin x 3 2. sinh x = | =>- cosh x = y 1 + sinh 2 x = -i/l sech x 16 9 ¥ = | , tanh x — 9 3 cosh x sinh x (3 ) § = 5> cothx =sk = i cosh x | , and csch x = ^4— = | 5 ' sinh x 4 3. cosh x = || , x > =>• sinh x = v cosh 2 x — 1 289 225 _64 225 ~~ 15 *■ , tanh x = ^i - i ' cosh x £ , coth x = -4— 17 tanh x sech x = — \— = || , and csch x = -r\— = ~ cosh x 17 ' sinh x a m 4. cosh x coth x 13 5 X > =>- sinh x = v cosh 2 *-* = Vf - 1 1 X 13 12 , sech x = l 5 " 13 , and csch x = l tanh cosh x sinh > vl - 1 (4 x +e -!„xN _ . phix , 1 -* + I 144 _ 12 sinh x , , tanhx — 25 5 cosh x 12 13 6. sinh (2 In x) e 21nx_ e -21„x tt2 1 I x "?J x 4 -l 2x2 7. cosh 5x + sinh 5x e 5x + e" 5x i e 5x - e" 5x Ox 8. cosh 3x — sinh 3x = e ~ + e ' — ^- L - 9. (sinh x + cosh x) 4 / e x — e x I e x +e x\4 4x (e x ) 4 = e' 10. In (cosh x + sinh x) + In (cosh x — sinh x) = In (cosh 2 x — sinh 2 x) = In 1 = 11. (a) sinh 2x = sinh (x + x) = sinh x cosh x + cosh x sinh x = 2 sinh x cosh x (b) cosh 2x = cosh (x + x) = cosh x cosh x + sinh x sin x = cosh 2 x + sinh 2 x 12. cosh 2 x - sinh 2 x = ("^f 1 ) 2 - ("^f 1 ) 2 = \ [(e x + e" x ) + (e x - e~ x )] [(e x + e" x ) - (e x - e~ x )] = \ (2e x ) (2e- x ) = \ (4e°) = \ (4) = 1 13. y = 6 sinh f => & = 6 (cosh f ) (±) = 2 cosh f 14. y = \ sinh(2x+ 1) %, = \ [cosh(2x + 1)](2) = cosh(2x + 1) 15. y = 20 tanh = 21 1 / 2 tanh t 1 / 2 =>. g = [sech 2 (t 1 / 2 )] (If 1 ^) (2tV 2 ) + (tanh t 1 / 2 ) (r 1 / 2 ) = sech 2 0+^^ 16. y = t 2 tanh \ = t 2 tanh t" 1 => % = [sech 2 (r 1 )] (-r 2 ) (t 2 ) + (2t) (tanh t -1 ) = - sech 2 \ + 2t tanh i 17. y = In (sinh z) =>■ & = ^ = coth z •' v 7 dz sinh z 18. y = In (cosh z) => & = ^4 = lanh z dz cosh z Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.8 Hyperbolic Functions 471 19. y = (sech 0)(1 - In sech 9) dy — sech 9 tanh i ! (sech 0) + (- sech tanh 0)(1 - In sech 0) sech tanh - (sech tanh 0)(l - In sech 0) = (sech tanh 0)[1 - (1 - In sech 0)] (sech tanh 0)(ln sech 0) 20. y = (csch 0)(1 - In csch 0) dy (csch 0) i - csch 8 coth \ (1 - In csch 0)(- csch coth 0) = csch coth - (1 - In csch 0)(csch coth 0) = (csch coth 0)(1 - 1 + In csch 0) = (csch coth 0)(ln csch 0) 21. y = In cosh v - \ tanh 2 v => ^ = ^^ - (|) (2 tanh v) (sech 2 v) = tanh v - (tanh v) (sech 2 v) = (tanh v) ( 1 — sech 2 v) = (tanh v) (tanh 2 v) = tanh 3 v 22. y = In sinh v - \ coth 2 v => ^ = f^ - (|) (2 coth v) (- csch 2 v) = coth v + (coth v) (csch 2 v) = (coth v) (1 + csch 2 v) = (coth v) (coth 2 v) = coth 3 v 23. y = (x 2 + 1) sech (In x) = (x 2 + 1) (^r^) = (x 2 + 1) (^) = (x 2 + 1) (^) = 2x =► g = 2 y = (4x 2 - 1) csch (In 2x) = (4x 2 - 1) ( e „, 2 , + 2 e _^ ) = (4x 2 - 1) (j^h^) = ^ ~ ! ) (sb) 24 4x =► & = 4 dx 25. y = sinh- 1 ^ = sinh" 1 (x 1 / 2 ) => £ = 2 ^ = ^ = -U ■ J V V / dx V 1+(x i /2) 2 2V^\/TT^ 2 x /x(l+x) 26. y = cosh" 1 2a/x + 1 = cosh" 1 (2(x + l) 1 / 2 1 dy= (2)(l)(x + l)-^ = 1 = l dx vW+D 1/2 ] 2 -l V^+lV^+3 \/4x2 + 7x + 3 27. y = (1 -0)tanh" 1 -1 a _*. & - (1 - ^) (i3p) + (-1) tanh_1 * = TTe ~ tanh_1 * 28. y = (0 2 + 20) tanh" 1 (0+1) => | = (0 2 + 20) [ 1 _ (9 1 +1)2 ] + (20 + 2) tanh" 1 (0 + 1) - -^±2»_ _l (261 + 2) tanh" 1 (0 + 1) = (20 + 2) tanh" 1 (0 + 1) - 1 -29 29. y = (1 - ^coth^v/t = (1 - Ocottr 1 (t 1 / 2 ) i=d-t) (i)'- i/2 1 - (tl/2) 2 + (-l)coth- 1 t 1 ft 1 / 21 ! 1 2\/t coth _1 \/t 30. y = (1 -t 2 )coth" 1 t => f = (1 -t 2 ) (y^) + (-2t) coth" 1 1 = 1 -2tcoth- 1 t 31. y = cos x x — x sech x x => = — sech -1 x dy _ -1 dx Vl-x 2 (—7==) + (1) sech" 1 xj = -j^= + -4— - sech" 1 x 32. y = In x + a/1 — x 2 sech : x = In x + (1 — x 2 ) sech 1 x => — = i + C 1 " x2 ) 1/2 (xTlb) + (J) C 1 " ^) _1/2 (-2x) sech- 1 x dx -1 „ _ 1 1 y/T- sech x v/T" sech x 33. y = csch" 1 (±) dy In (1)- In (2) _ In 2 i+4 i+4 Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 472 Chapter 7 Transcendental Functions ,-1 off -^ d Y _ On 2)2° _ -In 2 34. y = csch" 1 2" => 1 ^l + (2») 2 V^h - : 35. y = sinh x (tan x) => dy sec 2 x _ sec 2 x _ sec x| sec x sec X dx ^1 +(tanx) 2 y- sec^ x 36. y = cosh- 1 (sec x) =>■ £ = ( 7 x) , (tanx) = (sec , x)( ' an x) = (se " x)(ta " x) = sec x, 0< x < f J V ' dx \/sec 2 x - 1 \Aan 2 x I 1 ™ x l 2 37. (a) If y = tan" 1 (sinh x) + C, then f = , :° sh x = ^* = sec h x, which verifies the formula v ' J v 7 ' dx 1 + sinn^ x cosh z x (b) If y = sin -1 (tanh x) + C, then djr _ sech 2 x — _ sech^x _ secll w iji cn verifies the formula dx V 1 - tanh 2 x secn x 38. If y = f sech" 1 x - \ y/l - x 2 + C, then g = x sech" 1 x + f 2 J ^" * 2 which verifies the formula 2 f -1 1 | 2x 2 Vxv'l-x 2 / 4\/l -x 2 x sech : x, 39. If y = ^=1 com- 1 x + | + C, then -| = x coth -1 x + ( ^y 1 J (73^2) + 5 = * coth -1 x, which verifies the formula 40. If y = x tann -1 x + \ In (1 - x 2 ) + C, then -| = tanh -1 x + x (7^2) + 5 (7^2) = tanh -1 x, which verifies the formula 41. I sinh 2x dx = \ J sinh u du, where u = 2x and du = 2 dx cosh u + c cosh 2x + c 42. I sinh I dx = 5 / sinh u du, where u = | and du = i dx 5 cosh u + C = 5 cosh | + C 43. I 6 cosh (| — In 3) dx = 12 I cosh u du, where u = | — In 3 and du = | dx = 12 sinh u + C = 12 sinh (§ - In 3) + C 44. / 4 cosh (3x — In 2) dx = | I cosh u du, where u = 3x — In 2 and du = 3 dx | sinh u + C = 5 sinh (3x - In 2) + C 45. J tanh f dx = 7 J ^*jl du, where u = f and du = i dx = 7 In |cosh u| + Ci = 7 In |cosh || + Ci = 7 In = 7 1n|e x / 7 +e- x / 7 |+C /'+eW Ci = 7 In |e x / 7 + e- x / 7 | - 7 In 2 + Ci 46. f coth 4? d6> = \/3 f ^ du, where u = 4? and du J ,/3 V J smhu ' ,/3 V^ 75 e«/V3- e -9/\/3 I + Ci a/3 In |sinhu| + Ci = a/3 In sinh 4- + Ci = y/3 In , a/3 In |e 9 /v^ - e-^/v^l - a/3 In 2 + d = a/3 In le^/v^ - e- fl /^| + C 47. I sech 2 (x — |) dx = I sech 2 u du, where u = (x — |) and du = dx = tanh u + C = tanh (x - i) + C Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Section 7.8 Hyperbolic Functions 473 48. I csch 2 (5 — x) dx = — / csch 2 u du, where u = (5 — x) and du = — dx = -(- coth u) + C = coth u + C = coth (5 - x) + C 49. / sech y t tanh yt — dt = 2 I sech u tanh u du, where u = \J\ = t 1 / 2 and du : 2(- sech u) + C = -2 sech y^t + C dt 2^1 50. / cschdnt^cothdnt) ^ = J ^ Q CQth u ^ where u = h t and dll = f = — csch u + C = — csch (In t) + C 51. f° 4 cothxdx= r'' ,4 £2^dx= C*'* ± du = [In \u\}l 5 /, s = In \%\ - In ||| = In \% • \ I = In f , Jl„2 J In 2 sinhx J 3/4 u L I I J 3/4 181 141 18 31 2 ' where u = sinh x, du = cosh x dx, the lower limit is sinh (In 2) = e " ~ e = — ~- = | and the upper .. .. A-(k\ limit is sinh (In 4) 52. J o " tanh 2x dx = // ^ dx = \ J] I du = i [In |u|] 1 7/8 = ± [in (f ) - In l] = I In f , where u = cosh 2x, du = 2 sinh (2x) dx, the lower limit is cosh 0=1 and the upper limit is cosh (2 In 2) = cosh (In 4) -In 4 I p -ln4 53. r ,n2 2e coshede= r >n2 2t e ( e ^f^)dd= r' nl {e 2S +\)d9=\^+e} '" 2 J-ln4 J-ln4 \ 2 J J-ln4 v ' L 2 J-ln4 = (^ - In 2) - (^ - In 4^ = (| - In 2) - (^ - In 4) = ^ - In 2 + 2 In 2 = ^ + In 2 54. / o ,n2 4e-" sinh (9 d6» = J^e"" f^ 5 ^) d0 = 2 ^'"'(l - er 29 ) d9 = 2 \o + ^1 " = 2[(ln2+£=^) - (0+£)] =2(ln2+i-i) = 21n2+±-l=ln4-| 55. /"^cosh (tan 0) sec 2 (9 d(9 = J_',cosh u du = [sinh u] \ = sinh(l) - sinh(-l) = ( s ^ £=i ) - (^^f^) e — e x — e * +e e — e , where u = tan 9, du = sec 2 9 d9, the lower limit is tan I 1 and the upper limit is tan (f J = 1 56. I 2 sinh (sin 9) cos 9 d9 = 2 I sinh u du = 2 [cosh u] = 2(cosh 1 — cosh 0) = 2 ( - ■ - 1 e + e _1 — 2, where u = sin 9, du = cos 9 d6, the lower limit is sin = and the upper limit is sin (f ) = 1 57. J' c -^f^l dt = / o '" 2 cosh u du = [sinh u]^ 12 = sinh (In 2) - sinh(0) = e ' n2 - e ~'" 2 - u = In t, du = j dt, the lower limit is In 1 = and the upper limit is In 2 58. f * *"")/* dx = 16j^ cosh u du = 16 [sinh u] \ = 16(sinh 2 - sinh 1) = 16 = 8 (e 2 - e- 2 - limit is v 4 = 2 — i -1 1 y i = j , where (■=**)] (e 2 — e 2 — e + e x ) , where u = yfx = x 1 / 2 , du = | x 1//2 dx = 5=7- , the lower limit is \J 1 = 1 and the upper 59. /cosh 2 (| ) dx = J_" ln2 cosh 2 x + 1 dx = 1 J_° (cosh x + 1) dx = \ [sinh x + x] ^ ln2 Copyright (c| 1 Pen Etation, k, publishing as Pearson Addison-Wesle 474 Chapter 7 Transcendental Functions \ [(sinh + 0) - (sinh(- In 2) - In 2)] = \ [(0 + 0) - ( e "' n2 ~ e ' n2 - In 2)] = \ -2 In 2 I(l-I+ln2) 3 + I ln2= | +lnv ^ 60. / o ' n '°4 sinh 2 (I) dx = J o '°'°4 ( cosh 2 x ~' ) dx = 2j o ln10 (cosh x - 1) dx = 2 [sinh x- x]J = 2[(sinh(ln 10) - In 10) - (sinh - 0)] = e lnl ° - e-" ,lu - 2 In 10 = 10 - ^ - 2 In 10 = 9.9 - 2 In 10 1 In 10 Jo 61. sinh 63. tanh -1 /-5a In + 25 144 =m( 62. cosh" 1 (f (f)=m(! + yf r T) In 3 -i(_n = 1 in ( ±M) In 3 2 64. coth- 1 (|) Iln(^)=Iln9 = ln3 65. sech 1( l)=ln(i±Vg^) In 3 66. csdi" 1 (- -±\ = In ( - y^ ■ 4/3 /x/3) In (-V3 + 2) 67. < a > X 2^ y^ [sinh" 1 |] 'f = sinh" 1 a/3 - sinh = sinh" 1 a/3 (b) sinh" 1 a/3 = In (a/3 + a/3 + l) = In (a/3 + 2\ 68. (a) / 1/3 6 dx 2 / / t x o , where u = 3x, du = 3 dx, a = 1 Jo ^a 2 + u 2 \/l +9x 2 Jo y'a = [2 sinh- 1 u] = 2 (sinh" 1 1 - sinh" 1 0) = 2 sinh" 1 1 (b) 2 sinh" 1 1 = 2 In (l + a/I 2 + l) = 2 In (l + a/2~) -1 5 4 69. (a) J 4 y^U, dx = [coth- 1 x] L = coth- 1 2 - coth (b) coth" 1 2 - coth" 1 I = i [in 3 - In (fg)] = i In i 70. (a) J ~ r ^ 3 dx = [tanh- 1 x] / 2 = tanh" 1 \ - tanh" 1 = tanh" 1 \ l + (l/2) \ _ 1 1 — (1/2) ^ 2 (b) tanh- 1 i = iln(|^^;'l = i In 3 71. < a > /, 3/13 dx X 12/13 1/5 x\/l - 16x 2 J 4/5 u\/a 2 -u : [- sech" 1 ul/'f = - sech- 1 % + sech" 1 f , du „ , where u = 4x, du = 4 dx, a = 1 (b) - sech" 1 % + sech" 1 \ = - In / l + x/l- (12/13) 2 \ , / l + x/l-(4/5)2 \ ^ (12/13) J 1" m \^ (4/5) ^ In ( 13 + ^f IT¥ )+ln( 5 + x/25- 16 4 In (5±3) _ In (13±5) = In 2 -In I In (2 • § ) = In \ 72. « r^fe=t^ csch - L !| |] \ = - i (csch- 1 1 - csch" 1 \) = \ (csch- 1 | - csch" 1 l) (b) I (csch- 1 I - csch- 1 1) = I [in (2 + g) - In (l + a/2)] = I In (f±$) , cos x , dx = I , du = [sinh -1 ul n = sinh -1 — sinh -1 = 0, where u = sin x, du = cos x dx v/l+sin 2 x Jo v/l+u 2 1 JO (b) sinh" 1 - sinh" 1 = In (o + \/0 + l) - In (o + a/0 + l) = Copyright (c| 1 Pen Education, k, publishing as Pearson Addison-Wesle Section 7.8 Hyperbolic Functions 475 74. (a) /' dx 1 x\/l+(lnx) 2 ~ Jo / 1 I , ** u ^ , where u = In x, du = - dx, a = 1 Jo va 2 + u 2 x sinh 1 u] = sinh : 1 — sinh : = sinh * 1 (b) sinh" 1 1 - sinh" 1 = In (l + a/1 2 + l) - In (o + \/0 2 + l) = In (l + y/l\ 75. (a) Let E(x) = f(x)+ / ( - x) and O(x) = **>-*-*> . Then E(x) + O(x) = f(x) + f( ~ x) + f(x) ~ 2 f( ~ x) ._ 2fW _ f(x) AlsQj E( _ x) = f(-x) + f(-(-x)) = f(x) + f(-x) = E(x) ^ E(x) j s even ^ and 2 * vv . *„„v-, ^y ~j 2 2 f(-x) - f(-(-x)) _ _ f(x) - f(-x) 2 2 O(-x) = n x > ^ * x " = - IW 2 *> = - O(x) => O(x) is odd. Consequently, f(x) can be written as a sum of an even and an odd function. (b) f(x) f(x) + f(-x) because f(x)-f(-x) 2 ^v~««ov 2 2f(x) if f is even and f(x) f(x)-f(-x) because f(x) + f(-x) iff is odd. Thus, if f is even f(x) = ^ + and if f is odd, f(x) = + ^ 76. y = sinh l x =>- x = sinh y =>- x => 2x = e y - ^ =>• 2xe y = e 2y - 1 => e 2y - 2xe y -1=0 2x±y / 4x 2 + 4 =>• e y = x + v x 2 + 1 =>• sinh 1 x = y = ln(x+ y x 2 + 1 ) • Since e y > 0, we cannot choose e y = x — y x 2 + 1 because x — y x 2 + 1 < 0. 77. (a) v = A /^tanhf ,/* t ) =* dv _ /mg dt ~~ V k sech - 2 ./|k 2 ./|k t • Thus m^ = mg sech 2 ( y ^- t J = mg ( 1 — tanh 2 when t = 0. = g sech - mg — kv 2 . Also, since tanh x = when x = 0, v = (b) lim v= lim ,/2I tanh ( J^t ' t->oo t->oo v k ' v ' m ^ lim tanh K t — > oo mg /n / mg k U; — A/ I, ( C ) V 0555 ~ V i=f°° = m = 80 ^ w 178 89 ft/sec 78. (a) s(t) = a cos kt + b sin kt ds -ak sin kt + bk cos kt =4> ^f = — ak 2 cos kt — bk 2 sin kt -k 2 (a cos kt + b sin kt) = — k 2 s(t) =4> acceleration is proportional to s. The negative constant — k 2 implies that the acceleration is directed toward the origin. (b) s(t) = a cosh kt + b sinh kt ds ak sinh kt + bk cosh kt |f = ak 2 cosh kt + bk 2 sinh kt = k 2 (a cosh kt + b sinh kt) = k 2 s(t) =S> acceleration is proportional to s. The positive constant k 2 implies that the acceleration is directed away from the origin. 79. =>- y = I i l - dx + / , x ■ dx => y = sech 1 (x) - yl - x 2 + C; x = 1 and J x\/1 — x 2 J */ 1 _ x 2 Q- = -1 _|_ X _ _ _^^ = u x - -^ d " x\/l-x 2 \/l-x 2 " ^ ^ x\/l-x 2 "^ """ J \/l y = 0^C = 0=>y = sech" 1 (x) - v 7 ! - x 2 nb . 80. To find the length of the curve: y = i cosh ax =>• y' = sinh ax => L = / y 1 + (sinh ax) 2 dx ^*b . pb =>■ L = I cosh ax dx = [- sinh axl = - sinh ab. Then the area under the curve is A = | - cosh ax dx Jo LaJoa Ji a = [4 sinh ax] = 4 sinh ab = Q) Q sinh ab) which is the area of the rectangle of height 1 and length L as claimed. 81. V = 7T J (cosh 2 x - sinh 2 x) dx = tt J 1 dx = 2tt 82. V f 1 2tt Jo lnv/3 sech x dx = 2n ftanh x ln\/3 27T /3- 1A/3 '3 + 1/V3 Cf|t (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 476 Chapter 7 Transcendental Functions pin \/5 pin \/5 83. y = \ cosh 2x => y' = sinh 2x=^L=| yT+ (sinh 2x) 2 dx = I cosh 2x dx = [| sinh 2x] ln\/5 lnV5 1(^)] -JP-J) 84. (a) Let the point located at (cosh u, 0) be called T. Then A(u) = area of the triangle AOTP minus the area I pcoshu I under the curve y = \J x 2 — 1 from A to T => A(u) = \ cosh u sinh u I \J x 2 — 1 dx. (b) A(u) = \ cosh u sinh u J V x 2 — 1 dx =>■ A'(u) = | (cosh 2 u + sinh 2 u) — ( y cosh 2 u 1 J (sinh u) | cosh 2 u + ^ sinh 2 u — sinh 2 u = \ (cosh 2 u — sinh 2 u) (1) (c) A'(u) = \ => A(u) = I + C, and from part (a) we have A(0) = =>• C = =>• A(u) = § => u = 2A 85. y = 4 cosh f => 1 + (^V = 1 + sinh 2 (|) = cosh 2 (|) ; the surface area is S = J^" 27ry W 1 + f^V dx = 8tt P ' cosh 2 (f ) dx = 4tt f ' (l + cosh § ) dx = 4tt [x + 2 sinh §1 '° 81 J-I11I6 \4/ J-lnI6^ }> L 2 J —In 16 = 47r[(ln81 +2 sinh (^-)) - (-In 16 + 2 sinh (^f^))] =47r[ln(81 • 16) + 2 sinh (In 9) + 2 sinh (In 4)] = 4tt [In (9 - 4) 2 + (e ln9 - e" 1 " 9 ) + (e'» 4 - e" 1 " 4 )] = 4tt [2 In 36 + (9 - \) + (4 - \)] = 4tt (4 In 6 + f + f ) : 4tt (4 In 6 + 32Q 3 + 135 ) = 16tt In 6 + ^ 86. (a) y = cosh x => ds = ^/(dx) 2 + (dy) 2 = ^/(dx) 2 + (sinh 2 x) (dx) 2 = cosh x dx; M x = J ^ y ds = J^cosh x ds = J '* 2 cosh 2 x dx = /'"'(cosh 2x + 1) dx = [^^ + x] j" 2 == | (e ln4 - e" 1 " 4 ) + In 2 16 In r ln2 . r ln2 2;M = 2J VI +sinh 2 xdx = 2 J cosh x dx = 2 [sinh x]^' 1 " = e 1 " 2 1 3 Therefore, y (b) x = 0,y « 1.09 M* _ ({f+'" 2 ) _ 5 , In = I + 1j y > ar, d by symmetry x = 0. (-In 2, 1.25)' ,,(0,1.09} 0,8 0.6 0.4 0.2 ~3tr (ln2, 1.25) y = cosh x ) 02 6!4 0'6 — x 87. (a) y = S cosh (| x) =* tan = g = (S) [g sinh (g x)] = sinh (g x) (b) The tension at P is given by T cos </> = H =>■ T = H sec </> = H^/l + tan 2 . = H cosh (gx)=w(S) cosh (gx)=wy H v / l + (sinhgx)^ s = - sinh ax => sinh ax = as => ax = sinh : as =>■ x = - sinh J as; y = 1 cosh ax = 7 y cosh 2 = 1 Vsinh 2 ax + 1 = 1 v^VTT = A /s 2 + A ax (a) Since the cable is 32 ft long, s = 16 and x = 15. From Exercise 88, x = - sinh l as => 15a = sinh : 16a =4> sinh 15a = 16a. Cfigl (c) 1 Pea Education, Inc., publishing as Pearson Addison-Wesle Chapter 7 Practice Exercises 477 (b) The intersection is near (0.042, 0.672). y - 16a 0.61*3 (c) Newton's method indicates that at a rj 0.0417525 the curves y = 16a and y = sinh 15a intersect. (d) T = wy w (2 lb) (0^7325) » 47.90 lb (e) The sag is icosh(15a) - \ « 4.85 ft. 32 ' 30 ~28 26 22 20 -£5 =10 : 3~ y = (23.951 )cosh(0.4175 x) -5 — is — 1$ CHAPTER 7 PRACTICE EXERCISES 1. y = lOe"*/ 5 => I = (10) (- |) e""/ 5 = -2e-*/ 5 2. y = v/2e^ => | = (y/l) (y/l) e^ x = 2e^ 3 1 ■ y = i xe 1 „„*t 1 „4x __, dy 1 r / 1 M . a 4x/i\l 1 //|_43t\ „~4x _j_ 1 „4x 1 „4x „„4x [x (4e 4x ) + e 4x (l)] - i (4e 4x ) = xe 4x + ± e 4 4. y = x 2 e" 2 / x = xV 2x_1 => g = x 2 [(2x~ 2 ) e" 2x ~'] + e" 2x ~'(2x) = (2 + 2x)e" 2x ~' = 2e" 2 / x (l + x) 5. y = In (sin 2 9) => | = 2(sin g) 2 (c fl os e) = ^ = 2 cot -^ V / dff sin 2 sin 8 6. y = In (sec 2 dy 2(sec 0)(sec tan ( 2 ten 6 7. y = log dy j_ _j^ 1 2 dx In 2 I (*2\ I (ln2)x y = log 5 (3x - 7) In 5 dx \\n.5> V3x-7 7 (In 5)(3x-7) 9. y ^ = 8- 1 (ln8)(-l) = -8-'(ln8) J2i . dy _ Q2t 10. y = 9 21 => f = 9 2, (ln 9)(2) = 9 2, (2 In 9) 11. y = 5x 3 - 6 => | = 5(3.6)x 2 - 6 = 18x 2 - e 12. y = v /2x-^ => | = (v^)(->/2 cf-v^-O = -2xM-0 Copyright (c) 1 Pearson Etati, Inc., publishing as Pearson Addison-Wesle 478 Chapter 7 Transcendental Functions 13. y = (x + 2)*+ 2 =>■ lny = ln(x + 2)*+ 2 = (x + 2)ln(x + 2) => £ = (x + 2) (^) + (l)ln(x + 2) =* |=(x + 2)*+ 2 [ln(x + 2)+l] 14. y = 2(lnx)*/ 2 => In y = In [2(ln x)*/ 2 ] = In (2) + (|) In (In x) =► J = + (§) [£*] + (In (In x)) (±) =* y' = [21b + (I) ln ( ln x )] 2 ( ln x ) x/2 = ( ln x ) x/2 t ln ( ln x > + £] 15. y = sin : \f\ - u 2 = sin l (1 - u 2 ) 1 ' 2 =^ l n _,^/2 _^ dy _ i(l-uT' /2 (-2u) du ' l-[(l-u 2 ) 1/2 ] ==,0< u< 1 uyl — u 2 v 1 — u : 16. y = sin" 1 (-U = sin" 1 v" 1 / 2 =* | _|v"3/2 2 Vl-u 2 \/l-(l-u 2 ) lulVl-u 2 -^ ./l _ (v-i/2) 2 2vV 2 \/l-v-i 2v 3/2 Jl^i 2v 3 / 2 v/^T 2vVv - 1 17. y = ln(cos : x) => y' :os x V 1 — x 2 COS -1 X 18. y = z cos -1 z - Vt - z 2 = z cos -1 z - (1 - z 2 ) 1 ' 2 =4> ^ : — COS Z ^ dz yi _ z 2 (I)(l-z 2 )- 1/2 (-2z) 7T- 'l-z 2 19. y = t tan" 1 1 - (|) ln t ^ % = tan" 1 1 + t (^ 20. y = (1 + t 2 ) cor 1 2t => I = 2t cor 1 2t+ (1 + t 2 ) (y^ P )-a)(^)=tan- 1 t +T ^-l .-1,_ C,2 _ nV2 . dy 21. y = z sec : z — y z 2 — 1 = z sec : z — (z 2 — l) 1/z = + sec -1 z z 2 - 1 Vz 2 - 1 dz 1 ' ^- - 1 - sec -1 z, z > 1 ; (n^r^) + ( sec_1 z ) (D - \ ( z2 - i) _1/2 (2z) \ |z| v z — 1 / Vz 2 -1 22. y = 2a/x- 1 sec" 1 0c = 2(x - l) 1 / 2 sec- 1 fx 1 / dy _ 9 dx — ^ (!) (x - 1)-V 2 sec- 1 (x 1 / 2 ) + (x - l) 1 / 2 [ -Qfc^ xy x — 1 (sec * 27T 1 ■ 2x 7 x/x-i x 2 \ sec 23. y = esc 1 (sec 0) dy — sec 8 tan _ tan fl __ i C\ ^ f) ^ H |sec 0| ./sec 2 9 - 1 " l tan "I ~ ' 2 24. y = (1 + x 2 ) e tan x => y' = 2xe' _lx + (l + x 2 )(^5) =2xe tan " lx + tan L x I ~tan L x 25. y = ^^lny = ln(^)=ln(2) + ln(x 2 + l)-Iln(cos2x)^^=0+^ r -(i) '1^ (-2 sin 2x) cos 2x /'=(^T + tan 2x) y 2(x 2 + l) I 2x Vcos 2x Vx 2 + 1 tan 2x) 26 - y = yir^ => ln y = ln T^ = h M 3x + 4 ) - ln ( 2x - 4 )1 y^ - J_ ( 3 y 10 V3x + 4 2x-4V v ' _ J_ f_J L_\ v _ io/ 3x + 4 / 1 \ /_3 1.^ * 10 V3x + 4 x-2) J V 2x-4 VW V3x + 4 x-2-* Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle Chapter 7 Practice Exercises 479 27. y= [ fr|§^j ] 5 => hi y = 5 [ln(t + 1) + ln(t- 1) - ln(t - 2) - ln(t+ 3)] 9* ^ Vt+1 ~ t-1 t-2 t + 3/ dy g (t+l)(t-l) M, J l i_\ dt J I (t-2)(t + 3) I Vt+l T t-l t-2 t + 3/ 1 28 ' y = #h =* Iny = ln2 + lnu + uln2-lln(u 2 + l) => (l) (|) = I +ln2 - \ (^ \/u 2 +l dy = du x / u2 + 1 --^^^-^ 29. y = (sin0)^ =► In y = y/e In (sin 6) => (i) (|) = v^ (^|) + ± 0" 1 / 2 in (sin 0) => | = (sin 0)V" ( V^ cot 9 + In (sin 9) i /n 30. y = (lnx) 1 /inx ^ l n y = (^) ln(ln x) => £ = (£) (£) (I) + In (In x) L(]nx)2J uy ^y' = (lnx)^x[i_i^] 31. J e x sin (e x ) dx = J sin u du, where u = e x and du = e x dx = — cos u + C = — cos (e x ) + C 32. J e' cos (3e l — 2) dt = I J cos u du, where u = 3e l — 2 and du = 3e l dt = ± sin u + C = i sin (3e l - 2) + C 33. J e x sec 2 (e x — 7) dx = J sec 2 u du, where u = e x — 7 and du = e x dx = tan u + C = tan (e x - 7) + C 34. J e y esc (e y + 1) cot (e y + 1) dy = J esc u cot u du, where u = e y + 1 and du = e y dy = - esc u + C = - esc (e y + 1) + C 35. J (sec 2 x) e tanx dx = J e u du, where u = tan x and du = sec 2 x dx = e u + C = e tanx + C 36. J (esc 2 x) e cotx dx = — J e u du, where u = cot x and du = — esc 2 x dx = -e u + C = -e cotx + C 37. I ., : . dx = \ I - du, where u = 3x — 4, du = 3 dx; x = — 1 =>• u = —7, x = 1 =>■ u J_! 3x-4 3J_ 7 u = | [In |u|] Z\ = | [In |-1| - 1b |-7|] = I [0 - In 7] = - f 38. f ^P